Question

Source

• GeeksforGeeks: Find subarray with given sum - GeeksforGeeks

Given an nonnegative integer array, find a subarray where the sum of numbers is k.
Your code should return the index of the first number and the index of the last number.

Example
Given [1, 4, 20, 3, 10, 5], sum k = 33, return [2, 4].

题解 1 - 哈希表

题 Zero Sum Subarray | Data Structure and Algorithm 的升级版，这道题求子串和为 K 的索引。首先我们可以考虑使用时间复杂度相对较低的哈希表解决。前一道题的核心约束条件为 f(i1)−f(i2)=0f(i_1) - f(i_2) = 0f(i1)−f(i2)=0，这道题则变为 f(i1)−f(i2)=kf(i_1) - f(i_2) = kf(i1)−f(i2)=k

C++

#include <iostream>
#include <vector>
#include <map>

using namespace std;

class Solution {
public:
  /**
   * @param nums: A list of integers
   * @return: A list of integers includes the index of the first number
   *      and the index of the last number
   */
  vector<int> subarraySum(vector<int> nums, int k){
    vector<int> result;
    // curr_sum for the first item, index for the second item
    // unordered_map<int, int> hash;
    map<int, int> hash;
    hash[0] = 0;

    int curr_sum = 0;
    for (int i = 0; i != nums.size(); ++i) {
      curr_sum += nums[i];
      if (hash.find(curr_sum - k) != hash.end()) {
        result.push_back(hash[curr_sum - k]);
        result.push_back(i);
        return result;
      } else {
        hash[curr_sum] = i + 1;
      }
    }

    return result;
  }
};

int main(int argc, char *argv[])
{
  int int_array1[] = {1, 4, 20, 3, 10, 5};
  int int_array2[] = {1, 4, 0, 0, 3, 10, 5};
  vector<int> vec_array1;
  vector<int> vec_array2;
  for (int i = 0; i != sizeof(int_array1) / sizeof(int); ++i) {
    vec_array1.push_back(int_array1[i]);
  }
  for (int i = 0; i != sizeof(int_array2) / sizeof(int); ++i) {
    vec_array2.push_back(int_array2[i]);
  }

  Solution solution;
  vector<int> result1 = solution.subarraySum(vec_array1, 33);
  vector<int> result2 = solution.subarraySum(vec_array2, 7);

  cout << "result1 = [" << result1[0] << " ," << result1[1] << "]" << endl;
  cout << "result2 = [" << result2[0] << " ," << result2[1] << "]" << endl;

  return 0;
}

源码分析

与 Zero Sum Subarray 题的变化之处有两个地方，第一个是判断是否存在哈希表中时需要使用 hash.find(curr_sum - k) , 最终返回结果使用 result.push_back(hash[curr_sum - k]); 而不是 result.push_back(hash[curr_sum]);

复杂度分析

略，见 Zero Sum Subarray | Data Structure and Algorithm

题解 2 - 利用单调函数特性

不知道细心的你是否发现这道题的隐含条件—— nonnegative integer array , 这也就意味着子串和函数 f(i)f(i)f(i) 为「单调不减」函数。单调函数在数学中可是重点研究的对象，那么如何将这种单调性引入本题中呢？不妨设 i2>i1i_2 > i_1i2>i1, 题中的解等价于寻找 f(i2)−f(i1)=kf(i_2) - f(i_1) = kf(i2)−f(i1)=k, 则必有 f(i2)≥kf(i_2) \geq kf(i2)≥k.

我们首先来举个实际例子帮助分析，以整数数组 {1, 4, 20, 3, 10, 5} 为例，要求子串和为 33 的索引值。首先我们可以构建如下表所示的子串和 f(i)f(i)f(i).

f(i)f(i)f(i)	1	5	25	28	38
iii	0	1	2	3	4

要使部分子串和为 33，则要求的第二个索引值必大于等于 4，如果索引值再继续往后遍历，则所得的子串和必大于等于 38，进而可以推断出索引 0 一定不是解。那现在怎么办咧？当然是把它扔掉啊！第一个索引值往后递推，直至小于 33 时又往后递推第二个索引值，于是乎这种技巧又可以认为是「两根指针」。

C++

#include <iostream>
#include <vector>
#include <map>

using namespace std;

class Solution {
public:
  /**
   * @param nums: A list of integers
   * @return: A list of integers includes the index of the first number
   *      and the index of the last number
   */
  vector<int> subarraySum2(vector<int> &nums, int k){
    vector<int> result;

    int left_index = 0, curr_sum = 0;
    for (int i = 0; i != nums.size(); ++i) {
      while (curr_sum > k) {
        curr_sum -= nums[left_index];
        ++left_index;
      }

      if (curr_sum == k) {
        result.push_back(left_index);
        result.push_back(i - 1);
        return result;
      }
      curr_sum += nums[i];
    }
    return result;
  }
};

int main(int argc, char *argv[])
{
  int int_array1[] = {1, 4, 20, 3, 10, 5};
  int int_array2[] = {1, 4, 0, 0, 3, 10, 5};
  vector<int> vec_array1;
  vector<int> vec_array2;
  for (int i = 0; i != sizeof(int_array1) / sizeof(int); ++i) {
    vec_array1.push_back(int_array1[i]);
  }
  for (int i = 0; i != sizeof(int_array2) / sizeof(int); ++i) {
    vec_array2.push_back(int_array2[i]);
  }

  Solution solution;
  vector<int> result1 = solution.subarraySum2(vec_array1, 33);
  vector<int> result2 = solution.subarraySum2(vec_array2, 7);

  cout << "result1 = [" << result1[0] << " ," << result1[1] << "]" << endl;
  cout << "result2 = [" << result2[0] << " ," << result2[1] << "]" << endl;

  return 0;
}

源码分析

使用 for 循环，在 curr_sum > k 时使用 while 递减 curr_sum , 同时递增左边索引 left_index , 最后累加 curr_sum 。如果顺序不对就会出现 bug, 原因在于判断子串和是否满足条件时在递增之后（谢谢 @glbrtchen 汇报 bug)。

复杂度分析

看似有两重循环，由于仅遍历一次数组，且索引最多挪动和数组等长的次数。故最终时间复杂度近似为 O(2n)O(2n)O(2n), 空间复杂度为 O(1)O(1)O(1).

Reference

• Find subarray with given sum - GeeksforGeeks