Question

Source

• lintcode: (42) Maximum Subarray II

Given an array of integers,
find two non-overlapping subarrays which have the largest sum.

The number in each subarray should be contiguous.

Return the largest sum.

Example
For given [1, 3, -1, 2, -1, 2],
the two subarrays are [1, 3] and [2, -1, 2] or [1, 3, -1, 2] and [2],
they both have the largest sum 7.

Note
The subarray should contain at least one number

Challenge
Can you do it in time complexity O(n) ?

题解

严格来讲这道题这道题也可以不用动规来做，这里还是采用经典的动规解法。 Maximum Subarray 中要求的是数组中最大子数组和，这里是求不相重叠的两个子数组和的和最大值，做过买卖股票系列的题的话这道题就非常容易了，既然我们已经求出了单一子数组的最大和，那么我们使用隔板法将数组一分为二，分别求这两段的最大子数组和，求相加后的最大值即为最终结果。隔板前半部分的最大子数组和很容易求得，但是后半部分难道需要将索引从 0 开始依次计算吗？NO!!! 我们可以采用从后往前的方式进行遍历，这样时间复杂度就大大降低了。

Java

public class Solution {
  /**
   * @param nums: A list of integers
   * @return: An integer denotes the sum of max two non-overlapping subarrays
   */
  public int maxTwoSubArrays(ArrayList<Integer> nums) {
    // -1 is not proper for illegal input
    if (nums == null || nums.isEmpty()) return -1;

    int size = nums.size();
    // get max sub array forward
    int[] maxSubArrayF = new int[size];
    forwardTraversal(nums, maxSubArrayF);
    // get max sub array backward
    int[] maxSubArrayB = new int[size];
    backwardTraversal(nums, maxSubArrayB);
    // get maximum subarray by iteration
    int maxTwoSub = Integer.MIN_VALUE;
    for (int i = 0; i < size - 1; i++) {
      // non-overlapping
      maxTwoSub = Math.max(maxTwoSub, maxSubArrayF[i] + maxSubArrayB[i + 1]);
    }

    return maxTwoSub;
  }

  private void forwardTraversal(List<Integer> nums, int[] maxSubArray) {
    int sum = 0, minSum = 0, maxSub = Integer.MIN_VALUE;
    int size = nums.size();
    for (int i = 0; i < size; i++) {
      minSum = Math.min(minSum, sum);
      sum += nums.get(i);
      maxSub = Math.max(maxSub, sum - minSum);
      maxSubArray[i] = maxSub;
    }
  }

  private void backwardTraversal(List<Integer> nums, int[] maxSubArray) {
    int sum = 0, minSum = 0, maxSub = Integer.MIN_VALUE;
    int size = nums.size();
    for (int i = size - 1; i >= 0; i--) {
      minSum = Math.min(minSum, sum);
      sum += nums.get(i);
      maxSub = Math.max(maxSub, sum - minSum);
      maxSubArray[i] = maxSub;
    }
  }
}

源码分析

前向搜索和逆向搜索我们使用私有方法实现，可读性更高。注意是求非重叠子数组和，故求 maxTwoSub 时 i 的范围为 0, size - 2 , 前向数组索引为 i, 后向索引为 i + 1.

复杂度分析

前向和后向搜索求得最大子数组和，时间复杂度 O(2n)=O(n)O(2n)=O(n)O(2n)=O(n), 空间复杂度 O(n)O(n)O(n). 遍历子数组和的数组求最终两个子数组和的最大值，时间复杂度 O(n)O(n)O(n). 故总的时间复杂度为 O(n)O(n)O(n), 空间复杂度 O(n)O(n)O(n).