LeetCode 523. Continuous Subarray Sum

题目：
Given a list of non-negative numbers and a target integer k, write a function to check if the array has a continuous subarray of size at least 2 that sums up to the multiple of k, that is, sums up to n*k where n is also an integer.

Example 1:

Input: [23, 2, 4, 6, 7], k=6
Output: True
Explanation: Because [2, 4] is a continuous subarray of size 2 and sums up to 6.

Example 2:

Input: [23, 2, 6, 4, 7], k=6
Output: True
Explanation: Because [23, 2, 6, 4, 7] is an continuous subarray of size 5 and sums up to 42.

Note:

1. The length of the array won’t exceed 10,000.
2. You may assume the sum of all the numbers is in the range of a signed 32-bit integer.

思路： 利用动态规划的思想，计算nums中前i个元素的和，分别存到sum[i]中，这样能减少子问题的重复计算，然后再计算任意区间长度的值，如果k为0且有长度大于2的区间连续为0或者区间的和余k为0，则返回true。

代码：

class Solution {public:    bool checkSubarraySum(vector<int>& nums, int k) {        int size = nums.size();        if (size == 0){//如果nums为空，则返回false            return false;        }        vector<int> sum(size);        sum[0] = nums[0];//sum[i]中存的为nums中前i元素的和，这样能一定程度减少重复计算        for (int i = 1; i < size; ++i){            sum[i] = sum[i - 1] + nums[i];        }        for (int i = 0; i < size; ++i){            for (int j = i + 1; j < size; ++j){//计算了任意长度区间的nums元素和与k进行比较                if ((k == 0 && (sum[j] - sum[i] + nums[i]) == 0) || (k != 0 && (sum[j] - sum[i] + nums[i]) % k == 0)){                    return true;//如果k为0且有长度大于2的区间连续为0或者区间的和余k为0，则返回true                }            }        }        return false;//如果没有则返回false    }};

结果： 59ms