523. Continuous Subarray Sum Add to List
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.
1 2 3 4 5 6
1 3 6 10 15 21

思路：这里简单的记录一下当前子数组的和sum就可以了，注意一点当k是0的情况下的处理，如果k是0且数组中有连续的0，这种情况下也是可以的。

 1 bool checkSubarraySum(vector<int>& nums, int k)
 2 {
 3     int size = nums.size();
 4     if (size == 0)
 5         return false;
 6     vector<int>sum(size, 0);
 7     sum[0] = nums[0];
 8     for (int i = 1; i < size; i++)
 9         sum[i] = sum[i - 1] + nums[i];
10     for (int i = 0; i < size; i++)
11         for (int j = i + 1; j < size; j++)
12             if (k == 0 && (sum[j] - sum[i] + nums[i] ==0)||
13                 (k!=0&&(sum[j] - sum[i] + nums[i]) % k == 0))
14                 return true;
15     return false;
16 }

LeetCode:Continuous Subarray Sum