最大单笔卖出利润

Question

假设我们有一个由 n 个整数组成的数组，代表一天的股票价格。我们想要找到一对(buyDay, sellDay)，其中buyDay ≤ sellDay，这样如果我们在buyDay买入股票并卖出它在sellDay，我们将最大化我们的利润。

显然，该算法有一个 O(n²) 解决方案，方法是尝试所有可能的 (buyDay, sellDay) 对并找出最好的他们所有人。然而，是否有更好的算法，也许是在 O(n) 时间内运行的算法？

Answer 1

我喜欢这个问题。这是一个经典的面试问题，根据你如何思考，你最终会得到越来越好的解决方案。当然可以在比 O(n²) 更好的时间内完成此任务，并且我列出了三种不同的方式供您思考这个问题。

首先，分而治之的解决方案。让我们看看是否可以通过将输入分成两半，解决每个子数组中的问题，然后将两者组合在一起来解决这个问题。事实证明我们确实可以做到这一点，而且可以高效地做到这一点！直觉如下。如果我们只有一天，最好的选择是在那一天买入，然后在同一天卖回，没有利润。否则，将数组分成两半。如果我们考虑最佳答案可能是什么，它必须位于以下三个位置之一：

1. 正确的买入/卖出对完全发生在上半场。
2. 正确的买入/卖出对完全发生在下半场。
3. 正确的买入/卖出对发生在两个半场 - 我们在上半场买入，然后在下半场卖出。

我们可以通过在前半部分和后半部分递归调用我们的算法来获得 (1) 和 (2) 的值。对于方案（3），获得最高利润的方法是在上半年的最低点买入，在下半年的最高点卖出。只需对输入进行简单的线性扫描并找到两个值，我们就可以找到两半中的最小值和最大值。然后，这为我们提供了具有以下递归式的算法：

T(1) <= O(1)
T(n) <= 2T(n / 2) + O(n)

使用主定理来解决递归，我们发现它运行在 O(n lg n) 时间内，并且将使用 O(lg n) 空间进行递归调用。我们刚刚击败了简单的 O(n²) 解决方案！

但是等等！我们可以做得比这更好。请注意，我们在递归中使用 O(n) 项的唯一原因是我们必须扫描整个输入，试图找到每一半中的最小值和最大值。由于我们已经递归地探索了每一半，也许我们可以通过让递归也返回存储在每一半中的最小值和最大值来做得更好！换句话说，我们的递归返回三件事：

1. 最大化利润的买入和卖出时间。
2. 范围内总体的最小值。
3. 范围内的总体最大值。

最后两个值可以使用简单的递归来递归计算，我们可以在递归计算 (1) 的同时运行该递归：

1. 单个元素范围的最大值和最小值就是该元素。
2. 可以通过将输入分成两半，找到每一半的最大值和最小值，然后取它们各自的最大值和最小值来找到多元素范围的最大值和最小值。

如果我们使用这种方法，我们的递归关系现在是

T(1) <= O(1)
T(n) <= 2T(n / 2) + O(1)

使用主定理为我们提供了 O(n) 的运行时间和 O(lg n) 空间，这甚至比我们原来的解决方案更好！

但是等一下 - 我们可以做得更好！让我们考虑使用动态规划来解决这个问题。我们的想法是按如下方式思考这个问题。假设我们在查看前 k 个元素后就知道了问题的答案。我们能否利用我们对第 (k+1) 个元素的知识，结合我们的初始解决方案，来解决前 (k+1) 个元素的问题？如果是这样，我们可以通过解决第一个元素的问题，然后解决前两个，然后是前三个，等等，直到我们计算出前 n 个元素的问题来获得一个很好的算法。

让我们考虑一下如何做到这一点。如果我们只有一个元素，我们已经知道它一定是最佳的买入/卖出对。现在假设我们知道前 k 个元素的最佳答案，并查看第 (k+1) 个元素。那么，该值可以创建比前 k 个元素更好的解决方案的唯一方法是，前 k 个元素中最小的元素与该新元素之间的差异大于我们迄今为止计算的最大差异。因此，假设当我们遍历元素时，我们会跟踪两个值 - 迄今为止我们看到的最小值，以及仅使用前 k 个元素即可获得的最大利润。最初，我们到目前为止看到的最小值是第一个元素，最大利润为零。当我们看到一个新元素时，我们首先通过计算以迄今为止看到的最低价格购买并以当前价格出售可以赚多少钱来更新我们的最佳利润。如果这比我们迄今为止计算的最优值更好，那么我们将最优解更新为这个新的利润。接下来，我们将目前看到的最小元素更新为当前最小元素和新元素中的最小值。

由于在每个步骤中我们只执行 O(1) 工作，并且我们只访问 n 个元素中的每一个一次，因此需要 O(n) 时间才能完成！而且，它只使用O(1)辅助存储。这是我们迄今为止所取得的最好成果！

例如，根据您的输入，该算法的运行方式如下。数组每个值之间的数字对应于算法在该点保存的值。您实际上不会存储所有这些（这将占用 O(n) 内存！），但了解算法的演变很有帮助：

            5        10        4          6         7
min         5         5        4          4         4    
best      (5,5)     (5,10)   (5,10)     (5,10)    (5,10)

答案：(5, 10)

            5        10        4          6        12
min         5         5        4          4         4    
best      (5,5)     (5,10)   (5,10)     (5,10)    (4,12)

答案：(4, 12)

            1       2       3      4      5
min         1       1       1      1      1
best      (1,1)   (1,2)   (1,3)  (1,4)  (1,5)

答案：(1, 5 ）

我们现在可以做得更好吗？不幸的是，不是渐近意义上的。如果我们使用少于 O(n) 的时间，我们就无法查看大输入上的所有数字，因此不能保证我们不会错过最佳答案（我们可以将其“隐藏”在我们要查找的元素中）没看）。另外，我们不能使用小于 O(1) 的空间。可能会对隐藏在大 O 表示法中的常数因子进行一些优化，但除此之外，我们不能指望找到任何更好的选择。

总的来说，这意味着我们有以下算法：

• Naive：O(n²) 时间，O(1) 空间。
• 分而治之：时间 O(n lg n)，空间 O(lg n)。
• 优化的分而治之：O(n) 时间，O(lg n) 空间。
• 动态规划：时间 O(n)，空间 O(1)。

编辑：如果您有兴趣，我已经编写了这四种算法的 Python 版本，以便您可以使用它们并判断它们的相对性能。这是代码：

# Four different algorithms for solving the maximum single-sell profit problem,
# each of which have different time and space complexity.  This is one of my
# all-time favorite algorithms questions, since there are so many different
# answers that you can arrive at by thinking about the problem in slightly
# different ways.
#
# The maximum single-sell profit problem is defined as follows.  You are given
# an array of stock prices representing the value of some stock over time.
# Assuming that you are allowed to buy the stock exactly once and sell the
# stock exactly once, what is the maximum profit you can make?  For example,
# given the prices
#
#                        2, 7, 1, 8, 2, 8, 4, 5, 9, 0, 4, 5
#
# The maximum profit you can make is 8, by buying when the stock price is 1 and
# selling when the stock price is 9.  Note that while the greatest difference
# in the array is 9 (by subtracting 9 - 0), we cannot actually make a profit of
# 9 here because the stock price of 0 comes after the stock price of 9 (though
# if we wanted to lose a lot of money, buying high and selling low would be a
# great idea!)
#
# In the event that there's no profit to be made at all, we can always buy and
# sell on the same date.  For example, given these prices (which might
# represent a buggy-whip manufacturer:)
#
#                            9, 8, 7, 6, 5, 4, 3, 2, 1, 0
#
# The best profit we can make is 0 by buying and selling on the same day.
#
# Let's begin by writing the simplest and easiest algorithm we know of that
# can solve this problem - brute force.  We will just consider all O(n^2) pairs
# of values, and then pick the one with the highest net profit.  There are
# exactly n + (n - 1) + (n - 2) + ... + 1 = n(n + 1)/2 different pairs to pick
# from, so this algorithm will grow quadratically in the worst-case.  However,
# it uses only O(1) memory, which is a somewhat attractive feature.  Plus, if
# our first intuition for the problem gives a quadratic solution, we can be
# satisfied that if we don't come up with anything else, we can always have a
# polynomial-time solution.

def BruteForceSingleSellProfit(arr):
    # Store the best possible profit we can make; initially this is 0.
    bestProfit = 0;

    # Iterate across all pairs and find the best out of all of them.  As a
    # minor optimization, we don't consider any pair consisting of a single
    # element twice, since we already know that we get profit 0 from this.
    for i in range(0, len(arr)):
        for j in range (i + 1, len(arr)):
            bestProfit = max(bestProfit, arr[j] - arr[i])

    return bestProfit

# This solution is extremely inelegant, and it seems like there just *has* to
# be a better solution.  In fact, there are many better solutions, and we'll
# see three of them.
#
# The first insight comes if we try to solve this problem by using a divide-
# and-conquer strategy.  Let's consider what happens if we split the array into
# two (roughly equal) halves.  If we do so, then there are three possible
# options about where the best buy and sell times are:
#
# 1. We should buy and sell purely in the left half of the array.
# 2. We should buy and sell purely in the right half of the array.
# 3. We should buy in the left half of the array and sell in the right half of
#    the array.
#
# (Note that we don't need to consider selling in the left half of the array
# and buying in the right half of the array, since the buy time must always
# come before the sell time)
#
# If we want to solve this problem recursively, then we can get values for (1)
# and (2) by recursively invoking the algorithm on the left and right
# subarrays.  But what about (3)?  Well, if we want to maximize our profit, we
# should be buying at the lowest possible cost in the left half of the array
# and selling at the highest possible cost in the right half of the array.
# This gives a very elegant algorithm for solving this problem:
#
#    If the array has size 0 or size 1, the maximum profit is 0.
#    Otherwise:
#       Split the array in half.
#       Compute the maximum single-sell profit in the left array, call it L.
#       Compute the maximum single-sell profit in the right array, call it R.
#       Find the minimum of the first half of the array, call it Min
#       Find the maximum of the second half of the array, call it Max
#       Return the maximum of L, R, and Max - Min.
#
# Let's consider the time and space complexity of this algorithm.  Our base
# case takes O(1) time, and in our recursive step we make two recursive calls,
# one on each half of the array, and then does O(n) work to scan the array
# elements to find the minimum and maximum values.  This gives the recurrence
#
#    T(1) = O(1)
#    T(n) = 2T(n / 2) + O(n)
#
# Using the Master Theorem, this recurrence solves to O(n log n), which is
# asymptotically faster than our original approach!  However, we do pay a
# (slight) cost in memory usage.  Because we need to maintain space for all of
# the stack frames we use.  Since on each recursive call we cut the array size
# in half, the maximum number of recursive calls we can make is O(log n), so
# this algorithm uses O(n log n) time and O(log n) memory.

def DivideAndConquerSingleSellProfit(arr):
    # Base case: If the array has zero or one elements in it, the maximum
    # profit is 0.
    if len(arr) <= 1:
        return 0;

    # Cut the array into two roughly equal pieces.
    left  = arr[ : len(arr) / 2]
    right = arr[len(arr) / 2 : ]

    # Find the values for buying and selling purely in the left or purely in
    # the right.
    leftBest  = DivideAndConquerSingleSellProfit(left)
    rightBest = DivideAndConquerSingleSellProfit(right)

    # Compute the best profit for buying in the left and selling in the right.
    crossBest = max(right) - min(left)

    # Return the best of the three
    return max(leftBest, rightBest, crossBest)
    
# While the above algorithm for computing the maximum single-sell profit is
# better timewise than what we started with (O(n log n) versus O(n^2)), we can
# still improve the time performance.  In particular, recall our recurrence
# relation:
#
#    T(1) = O(1)
#    T(n) = 2T(n / 2) + O(n)
#
# Here, the O(n) term in the T(n) case comes from the work being done to find
# the maximum and minimum values in the right and left halves of the array,
# respectively.  If we could find these values faster than what we're doing
# right now, we could potentially decrease the function's runtime.
#
# The key observation here is that we can compute the minimum and maximum
# values of an array using a divide-and-conquer approach.  Specifically:
#
#    If the array has just one element, it is the minimum and maximum value.
#    Otherwise:
#       Split the array in half.
#       Find the minimum and maximum values from the left and right halves.
#       Return the minimum and maximum of these two values.
#
# Notice that our base case does only O(1) work, and our recursive case manages
# to do only O(1) work in addition to the recursive calls.  This gives us the
# recurrence relation
#
#    T(1) = O(1)
#    T(n) = 2T(n / 2) + O(1)
#
# Using the Master Theorem, this solves to O(n).
#
# How can we make use of this result?  Well, in our current divide-and-conquer
# solution, we split the array in half anyway to find the maximum profit we
# could make in the left and right subarrays.  Could we have those recursive
# calls also hand back the maximum and minimum values of the respective arrays?
# If so, we could rewrite our solution as follows:
#
#    If the array has size 1, the maximum profit is zero and the maximum and
#       minimum values are the single array element.
#    Otherwise:
#       Split the array in half.
#       Compute the maximum single-sell profit in the left array, call it L.
#       Compute the maximum single-sell profit in the right array, call it R.
#       Let Min be the minimum value in the left array, which we got from our
#           first recursive call.
#       Let Max be the maximum value in the right array, which we got from our
#           second recursive call.
#       Return the maximum of L, R, and Max - Min for the maximum single-sell
#           profit, and the appropriate maximum and minimum values found from
#           the recursive calls.
#
# The correctness proof for this algorithm works just as it did before, but now
# we never actually do a scan of the array at each step.  In fact, we do only
# O(1) work at each level.  This gives a new recurrence
#
#     T(1) = O(1)
#     T(n) = 2T(n / 2) + O(1)
#
# Which solves to O(n).  We're now using O(n) time and O(log n) memory, which
# is asymptotically faster than before!
#
# The code for this is given below:

def OptimizedDivideAndConquerSingleSellProfit(arr):
    # If the array is empty, the maximum profit is zero.
    if len(arr) == 0:
        return 0

    # This recursive helper function implements the above recurrence.  It
    # returns a triple of (max profit, min array value, max array value).  For
    # efficiency reasons, we always reuse the array and specify the bounds as
    # [lhs, rhs]
    def Recursion(arr, lhs, rhs):
        # If the array has just one element, we return that the profit is zero
        # but the minimum and maximum values are just that array value.
        if lhs == rhs:
            return (0, arr[lhs], arr[rhs])

        # Recursively compute the values for the first and latter half of the
        # array.  To do this, we need to split the array in half.  The line
        # below accomplishes this in a way that, if ported to other languages,
        # cannot result in an integer overflow.
        mid = lhs + (rhs - lhs) / 2
        
        # Perform the recursion.
        ( leftProfit,  leftMin,  leftMax) = Recursion(arr, lhs, mid)
        (rightProfit, rightMin, rightMax) = Recursion(arr, mid + 1, rhs)

        # Our result is the maximum possible profit, the minimum of the two
        # minima we've found (since the minimum of these two values gives the
        # minimum of the overall array), and the maximum of the two maxima.
        maxProfit = max(leftProfit, rightProfit, rightMax - leftMin)
        return (maxProfit, min(leftMin, rightMin), max(leftMax, rightMax))

    # Using our recursive helper function, compute the resulting value.
    profit, _, _ = Recursion(arr, 0, len(arr) - 1)
    return profit

# At this point we've traded our O(n^2)-time, O(1)-space solution for an O(n)-
# time, O(log n) space solution.  But can we do better than this?
#
# To find a better algorithm, we'll need to switch our line of reasoning.
# Rather than using divide-and-conquer, let's see what happens if we use
# dynamic programming.  In particular, let's think about the following problem.
# If we knew the maximum single-sell profit that we could get in just the first
# k array elements, could we use this information to determine what the
# maximum single-sell profit would be in the first k + 1 array elements?  If we
# could do this, we could use the following algorithm:
#
#   Find the maximum single-sell profit to be made in the first 1 elements.
#   For i = 2 to n:
#      Compute the maximum single-sell profit using the first i elements.
#
# How might we do this?  One intuition is as follows.  Suppose that we know the
# maximum single-sell profit of the first k elements.  If we look at k + 1
# elements, then either the maximum profit we could make by buying and selling
# within the first k elements (in which case nothing changes), or we're
# supposed to sell at the (k + 1)st price.  If we wanted to sell at this price
# for a maximum profit, then we would want to do so by buying at the lowest of
# the first k + 1 prices, then selling at the (k + 1)st price.
#
# To accomplish this, suppose that we keep track of the minimum value in the
# first k elements, along with the maximum profit we could make in the first
# k elements.  Upon seeing the (k + 1)st element, we update what the current
# minimum value is, then update what the maximum profit we can make is by
# seeing whether the difference between the (k + 1)st element and the new
# minimum value is.  Note that it doesn't matter what order we do this in; if
# the (k + 1)st element is the smallest element so far, there's no possible way
# that we could increase our profit by selling at that point.
#
# To finish up this algorithm, we should note that given just the first price,
# the maximum possible profit is 0.
#
# This gives the following simple and elegant algorithm for the maximum single-
# sell profit problem:
#
#   Let profit = 0.
#   Let min = arr[0]
#   For k = 1 to length(arr):
#       If arr[k] < min, set min = arr[k]
#       If profit < arr[k] - min, set profit = arr[k] - min
#
# This is short, sweet, and uses only O(n) time and O(1) memory.  The beauty of
# this solution is that we are quite naturally led there by thinking about how
# to update our answer to the problem in response to seeing some new element.
# In fact, we could consider implementing this algorithm as a streaming
# algorithm, where at each point in time we maintain the maximum possible
# profit and then update our answer every time new data becomes available.
#
# The final version of this algorithm is shown here:

def DynamicProgrammingSingleSellProfit(arr):
    # If the array is empty, we cannot make a profit.
    if len(arr) == 0:
        return 0

    # Otherwise, keep track of the best possible profit and the lowest value
    # seen so far.
    profit = 0
    cheapest = arr[0]

    # Iterate across the array, updating our answer as we go according to the
    # above pseudocode.
    for i in range(1, len(arr)):
        # Update the minimum value to be the lower of the existing minimum and
        # the new minimum.
        cheapest = min(cheapest, arr[i])

        # Update the maximum profit to be the larger of the old profit and the
        # profit made by buying at the lowest value and selling at the current
        # price.
        profit = max(profit, arr[i] - cheapest)

    return profit

# To summarize our algorithms, we have seen
#
# Naive:                        O(n ^ 2)   time, O(1)     space
# Divide-and-conquer:           O(n log n) time, O(log n) space
# Optimized divide-and-conquer: O(n)       time, O(log n) space
# Dynamic programming:          O(n)       time, O(1)     space

Answer 2

这是带有一点间接性的最大和子序列问题。最大和子序列问题给定一个可以是正数或负数的整数列表，找到该列表的连续子集的最大和。

您可以通过计算连续几天之间的利润或损失，轻松地将这个问题转换为另一个问题。因此，您可以将股票价格列表（例如 [5, 6, 7, 4, 2]）转换为收益/损失列表，例如 [1, 1, -3, -2]。那么子序列和问题就很容易解决： 查找数组中元素和最大的子序列

Answer 3

我不太确定为什么这被认为是动态规划问题。我在教科书和算法指南中看到过这个问题，使用 O(n log n) 运行时间和 O(log n) 空间（例如《编程面试要素》）。这似乎是一个比人们想象的要简单得多的问题。

这是通过跟踪最大利润、最低购买价格以及最佳购买/出售价格来实现的。当它遍历数组中的每个元素时，它会检查给定元素是否小于最低购买价格。如果是，则最低购买价格指数 (min) 将更新为该元素的索引。此外，对于每个元素，becomeABillionaire 算法会检查 arr[i] - arr[min]（当前元素与最低购买价格之间的差值）是否大于当前利润。如果是，则利润将更新为该差额，并将买入设置为 arr[min]，将卖出设置为 arr[i]。

单次运行。

static void becomeABillionaire(int arr[]) {
    int i = 0, buy = 0, sell = 0, min = 0, profit = 0;

    for (i = 0; i < arr.length; i++) {
        if (arr[i] < arr[min])
            min = i;
        else if (arr[i] - arr[min] > profit) {
            buy = min; 
            sell = i;
            profit = arr[i] - arr[min];
        }

    }

    System.out.println("We will buy at : " + arr[buy] + " sell at " + arr[sell] + 
            " and become billionaires worth " + profit );

}

合著者：https://stackoverflow.com/users/599402/ephraim

Answer 4

该问题与最大子序列相同
我用动态规划解决了这个问题。跟踪当前和以前（利润、买入日期和卖出日期）
如果当前值高于先前值，则用当前值替换先前值。

    int prices[] = { 38, 37, 35, 31, 20, 24, 35, 21, 24, 21, 23, 20, 23, 25, 27 };

    int buyDate = 0, tempbuyDate = 0;
    int sellDate = 0, tempsellDate = 0; 

    int profit = 0, tempProfit =0;
    int i ,x = prices.length;
    int previousDayPrice = prices[0], currentDayprice=0;

    for(i=1 ; i<x; i++ ) {

        currentDayprice = prices[i];

        if(currentDayprice > previousDayPrice ) {  // price went up

            tempProfit = tempProfit + currentDayprice - previousDayPrice;
            tempsellDate = i;
        }
        else { // price went down 

            if(tempProfit>profit) { // check if the current Profit is higher than previous profit....

                profit = tempProfit;
                sellDate = tempsellDate;
                buyDate = tempbuyDate;
            } 
                                     // re-intialized buy&sell date, profit....
                tempsellDate = i;
                tempbuyDate = i;
                tempProfit =0;
        }
        previousDayPrice = currentDayprice;
    }

    // if the profit is highest till the last date....
    if(tempProfit>profit) {
        System.out.println("buydate " + tempbuyDate + " selldate " + tempsellDate + " profit " + tempProfit );
    }
    else {
        System.out.println("buydate " + buyDate + " selldate " + sellDate + " profit " + profit );
    }   

Answer 5

这是我的Java解决方案：

public static void main(String[] args) {
    int A[] = {5,10,4,6,12};

    int min = A[0]; // Lets assume first element is minimum
    int maxProfit = 0; // 0 profit, if we buy & sell on same day.
    int profit = 0;
    int minIndex = 0; // Index of buy date
    int maxIndex = 0; // Index of sell date

    //Run the loop from next element
    for (int i = 1; i < A.length; i++) {
        //Keep track of minimum buy price & index
        if (A[i] < min) {
            min = A[i];
            minIndex = i;
        }
        profit = A[i] - min;
        //If new profit is more than previous profit, keep it and update the max index
        if (profit > maxProfit) {
            maxProfit = profit;
            maxIndex = i;
        }
    }
    System.out.println("maxProfit is "+maxProfit);
    System.out.println("minIndex is "+minIndex);
    System.out.println("maxIndex is "+maxIndex);     
}

Answer 6

我想出了一个简单的解决方案 - 代码更不言自明。这是动态规划问题之一。

该代码不考虑错误检查和边缘情况。它只是一个示例，给出解决问题的基本逻辑思路。

namespace MaxProfitForSharePrice
{
    class MaxProfitForSharePrice
    {
        private static int findMax(int a, int b)
        {
            return a > b ? a : b;
        }

        private static void GetMaxProfit(int[] sharePrices)
        {
            int minSharePrice = sharePrices[0], maxSharePrice = 0, MaxProft = 0;
            int shareBuyValue = sharePrices[0], shareSellValue = sharePrices[0];

            for (int i = 0; i < sharePrices.Length; i++)
            {
                if (sharePrices[i] < minSharePrice )
                {
                    minSharePrice = sharePrices[i];
                    // if we update the min value of share, we need to reset the Max value as 
                    // we can only do this transaction in-sequence. We need to buy first and then only we can sell.
                    maxSharePrice = 0; 
                }
                else 
                {
                    maxSharePrice = sharePrices[i];
                }

                // We are checking if max and min share value of stock are going to
                // give us better profit compare to the previously stored one, then store those share values.
                if (MaxProft < (maxSharePrice - minSharePrice))
                {
                    shareBuyValue = minSharePrice;
                    shareSellValue = maxSharePrice;
                }

                MaxProft = findMax(MaxProft, maxSharePrice - minSharePrice);
            }

            Console.WriteLine("Buy stock at ${0} and sell at ${1}, maximum profit can be earned ${2}.", shareBuyValue, shareSellValue, MaxProft);
        }

        static void Main(string[] args)
        {
           int[] sampleArray = new int[] { 1, 3, 4, 1, 1, 2, 11 };
           GetMaxProfit(sampleArray);
            Console.ReadLine();
        }
    }
}

Answer 7

public static double maxProfit(double [] stockPrices)
    {
        double initIndex = 0, finalIndex = 0;

        double tempProfit = list[1] - list[0];
        double maxSum = tempProfit;
        double maxEndPoint = tempProfit;


        for(int i = 1 ;i<list.length;i++)
        {
            tempProfit = list[ i ] - list[i - 1];;

            if(maxEndPoint < 0)
            {
                maxEndPoint = tempProfit;
                initIndex = i;
            }
            else
            {
                maxEndPoint += tempProfit;
            }

            if(maxSum <= maxEndPoint)
            {
                maxSum = maxEndPoint ;
                finalIndex = i;
            }
        }
        System.out.println(initIndex + " " + finalIndex);
        return maxSum;

    }

这是我的解决方案。修改最大子序列算法。在 O(n) 内解决问题。我认为这不能做得更快了。

Answer 8

这是一个有趣的问题，因为它看起来很难，但仔细思考会产生一个优雅的、简化的解决方案。

正如已经指出的，它可以在 O(N^2) 时间内暴力破解。对于数组（或列表）中的每个条目，迭代所有先前的条目以获取最小值或最大值，具体取决于问题是找到最大收益还是最大损失。

以下是如何考虑 O(N) 的解决方案：每个条目代表一个新的可能的最大值（或最小值）。然后，我们需要做的就是保存先前的最小值（或最大值），并将差异与当前和先前的最小值（或最大值）进行比较。简单易行。

以下是 Java 中的 JUnit 测试代码：

import org.junit.Test;

public class MaxDiffOverSeriesProblem {

    @Test
    public void test1() {
        int[] testArr = new int[]{100, 80, 70, 65, 95, 120, 150, 75, 95, 100, 110, 120, 90, 80, 85, 90};

        System.out.println("maxLoss: " + calculateMaxLossOverSeries(testArr) + ", maxGain: " + calculateMaxGainOverSeries(testArr));
    }

    private int calculateMaxLossOverSeries(int[] arr) {
        int maxLoss = 0;

        int idxMax = 0;
        for (int i = 0; i < arr.length; i++) {
            if (arr[i] > arr[idxMax]) {
                idxMax = i;
            }

            if (arr[idxMax] - arr[i] > maxLoss) {
                maxLoss = arr[idxMax] - arr[i];
            }           
        }

        return maxLoss;
    }

    private int calculateMaxGainOverSeries(int[] arr) {
        int maxGain = 0;

        int idxMin = 0;
        for (int i = 0; i < arr.length; i++) {
            if (arr[i] < arr[idxMin]) {
                idxMin = i;
            }

            if (arr[i] - arr[idxMin] > maxGain) {
                maxGain = arr[i] - arr[idxMin];
            }           
        }

        return maxGain;
    }

}

在计算最大损失的情况下，我们跟踪列表中的最大值（买入价格）直至当前条目。然后我们计算最大值和当前条目之间的差异。如果最大-电流> maxLoss，然后我们将此差异保留为新的 maxLoss。由于 max 的索引保证小于 current 的索引，因此我们保证“买入”日期小于“卖出”日期。

在计算最大增益的情况下，一切都相反。我们跟踪列表中直到当前条目的最小值。我们计算最小值和当前条目之间的差异（反转减法中的顺序）。如果电流-最小值> maxGain，然后我们将这个差异保留为新的 maxGain。同样，“买入”（最小值）的索引位于当前（“卖出”）的索引之前。

我们只需要跟踪 maxGain（或 maxLoss）以及 min 或 max 的索引，但不需要同时跟踪两者，并且我们不需要比较索引来验证“买入”小于“卖出”，因为我们自然地得到这个。

Answer 9

最大单次销售利润，O(n) 解决方案

function stocks_n(price_list){
    var maxDif=0, min=price_list[0]

    for (var i in price_list){
        p = price_list[i];
        if (p<min)
            min=p
        else if (p-min>maxDif)
                maxDif=p-min;
   }

    return maxDif
}

这是一个项目，它对 100k 整数的随机数据集上的 o(N) 与 o(n^2) 方法进行时间复杂度测试。 O(n^2) 需要 2 秒，而 O(n) 需要 0.01 秒

https://github.com/gulakov/complexity.js

function stocks_n2(ps){
    for (maxDif=0,i=_i=0;p=ps[i++];i=_i++)
        for (;p2=ps[i++];)
            if (p2-p>maxDif)
                maxDif=p2-p
    return maxDif
}

这是较慢的 o(n^2) 方法，每天循环遍历其余天，双循环。

Answer 10

得票最多的答案不允许最大利润为负的情况，应进行修改以允许此类情况。可以通过将循环范围限制为 (len(a) - 1) 并通过将指数移动 1 来改变确定利润的方式来实现此目的。

def singSellProfit(a):
profit = -max(a)
low = a[0]

for i in range(len(a) - 1):
    low = min(low, a[i])
    profit = max(profit, a[i + 1] - low)
return profit

将此版本的函数与之前的数组进行比较：

s = [19,11,10,8,5,2]

singSellProfit(s)
-1

DynamicProgrammingSingleSellProfit(s)
0

Answer 11

static void findmaxprofit(int[] stockvalues){
    int buy=0,sell=0,buyingpoint=0,sellingpoint=0,profit=0,currentprofit=0;
    int finalbuy=0,finalsell=0;
    if(stockvalues.length!=0){
        buy=stockvalues[0];
    }           
    for(int i=1;i<stockvalues.length;i++){  
        if(stockvalues[i]<buy&&i!=stockvalues.length-1){                
            buy=stockvalues[i];
            buyingpoint=i;
        }               
        else if(stockvalues[i]>buy){                
            sell=stockvalues[i];
            sellingpoint=i;
        }
        currentprofit=sell-buy;         
        if(profit<currentprofit&&sellingpoint>buyingpoint){             
            finalbuy=buy;
            finalsell=sell;
            profit=currentprofit;
        }

    }
    if(profit>0)
    System.out.println("Buy shares at "+finalbuy+" INR and Sell Shares "+finalsell+" INR and Profit of "+profit+" INR");
    else
        System.out.println("Don't do Share transacations today");
}

Answer 12

确定最大利润的一种可能性可能是跟踪数组中每个索引处的左侧最小元素和右侧最大元素。然后，当您迭代股票价格时，对于任何给定的一天，您都会知道截至该天的最低价格，并且您还会知道该天之后（包括该天）的最高价格。

例如，让我们定义一个 min_arr 和 max_arr，给定的数组为 arr。 min_arr 中的索引 i 将是所有索引 <= i 中 arr 中的最小元素（和 的左侧）包括我）。 max_arr 中的索引 i 将是所有索引 >= i 中 arr 中的最大元素（和 的右侧）包括我）。然后，您可以找到 max_arr 和 min_arr 中相应元素之间的最大差异：

def max_profit(arr)
   min_arr = []
   min_el = arr.first
   arr.each do |el|
       if el < min_el
           min_el = el
           min_arr << min_el
       else
           min_arr << min_el
       end
   end

   max_arr = []
   max_el = arr.last
   arr.reverse.each do |el|
       if el > max_el
           max_el = el
           max_arr.unshift(max_el)
       else
           max_arr.unshift(max_el)
       end

   end

   max_difference = max_arr.first - min_arr.first
   1.upto(arr.length-1) do |i|
        max_difference = max_arr[i] - min_arr[i] if max_difference < max_arr[i] - min_arr[i]  
   end

   return max_difference 
end

这应该在 O(n) 时间内运行，但我相信它会占用大量空间。

Answer 13

这是数组中两个元素之间的最大差异，这是我的解决方案：

O(N) 时间复杂度
O(1) 空间复杂度

    int[] arr   =   {5, 4, 6 ,7 ,6 ,3 ,2, 5};

    int start   =   0;
    int end     =   0;
    int max     =   0;
    for(int i=1; i<arr.length; i++){
        int currMax =   arr[i] - arr[i-1];
        if(currMax>0){
            if((arr[i] -arr[start])>=currMax && ((arr[i] -arr[start])>=(arr[end] -arr[start]))){

                 end    =   i;
            }
            else if(currMax>(arr[i] -arr[start]) && currMax >(arr[end] - arr[start])){
                start   =   i-1;
                end =   i;
            }
        }
    }
    max =   arr[end] - arr[start];
    System.out.println("max: "+max+" start: "+start+" end: "+end);

Answer 14

在 FB 解决方案工程师职位的现场编码考试中失败后，我必须在冷静的氛围中解决这个问题，所以这是我的 2 美分：

var max_profit = 0;
var stockPrices = [23,40,21,67,1,50,22,38,2,62];

var currentBestBuy = 0; 
var currentBestSell = 0;
var min = 0;

for(var i = 0;i < (stockPrices.length - 1) ; i++){
    if(( stockPrices[i + 1] - stockPrices[currentBestBuy] > max_profit) ){
        max_profit = stockPrices[i + 1] - stockPrices[currentBestBuy];
        currentBestSell = i + 1;  
    }
    if(stockPrices[i] < stockPrices[currentBestBuy]){
            min = i;
        }
    if( max_profit < stockPrices[i + 1] - stockPrices[min] ){
        max_profit = stockPrices[i + 1] - stockPrices[min];
        currentBestSell = i + 1;
        currentBestBuy = min;
    }
}

console.log(currentBestBuy);
console.log(currentBestSell);
console.log(max_profit);

Answer 15

真正回答问题的唯一答案是 @akash_magoon 的答案（并且以如此简单的方式！），但它不会返回问题中指定的确切对象。我重构了一点，并在 PHP 中得到了我的答案，返回了所要求的内容：

function maximizeProfit(array $dailyPrices)
{
    $buyDay = $sellDay = $cheaperDay = $profit = 0;

    for ($today = 0; $today < count($dailyPrices); $today++) {
        if ($dailyPrices[$today] < $dailyPrices[$cheaperDay]) {
            $cheaperDay = $today;
        } elseif ($dailyPrices[$today] - $dailyPrices[$cheaperDay] > $profit) {
            $buyDay  = $cheaperDay;
            $sellDay = $today;
            $profit   = $dailyPrices[$today] - $dailyPrices[$cheaperDay];
        }
    }
    return [$buyDay, $sellDay];
}

Answer 16

一个巧妙的解决方案：

+ (int)maxProfit:(NSArray *)prices {
    int maxProfit = 0;

    int bestBuy = 0;
    int bestSell = 0;
    int currentBestBuy = 0;

    for (int i= 1; i < prices.count; i++) {
        int todayPrice = [prices[i] intValue];
        int bestBuyPrice = [prices[currentBestBuy] intValue];
        if (todayPrice < bestBuyPrice) {
            currentBestBuy = i;
            bestBuyPrice = todayPrice;
        }

        if (maxProfit < (todayPrice - bestBuyPrice)) {
            bestSell = i;
            bestBuy = currentBestBuy;
            maxProfit = (todayPrice - bestBuyPrice);
        }
    }

    NSLog(@"Buy Day : %d", bestBuy);
    NSLog(@"Sell Day : %d", bestSell);

    return maxProfit;
}

Answer 17

def get_max_profit(stock):
    p=stock[0]
    max_profit=0
    maxp=p
    minp=p
    for i in range(1,len(stock)):
        p=min(p,stock[i])
        profit=stock[i]-p
        if profit>max_profit:
            maxp=stock[i]
            minp=p
            max_profit=profit
    return minp,maxp,max_profit



stock_prices = [310,315,275,295,260,270,290,230,255,250]
print(get_max_profit(stock_prices))

这个Python3程序可以返回使利润最大化的买入价和卖出价，计算的时间复杂度为O(n)，空间复杂度为O(1)。

Answer 18

这是我的解决方案

public static int maxProfit(List<Integer> in) {
    int min = in.get(0), max = 0;
    for(int i=0; i<in.size()-1;i++){

        min=Math.min(min, in.get(i));

        max = Math.max(in.get(i) - min, max);
     }

     return max;
 }
}

Answer 19

对于多次买入/卖出，
使用下面的代码

• 时间复杂度 O(n)

  n=int(input())
  a = 列表(map(int,input().split())) 
  米=0
  b=a[0]
  s=0
  对于范围 (1,n) 内的 i：
    if(a[i]b):
    m=m+a[n-1]-b
  打印(米)
  

Answer 20

对于所有跟踪最小和最大元素的答案，该解决方案实际上是一个 O(n^2) 解决方案。这是因为最后必须检查最大值是否出现在最小值之后。如果没有，则需要进一步迭代，直到满足该条件，这会留下 O(n^2) 的最坏情况。如果您想跳过额外的迭代，则需要更多的空间。不管怎样，与动态规划解决方案相比，这是一个禁忌