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=6Output: TrueExplanation: Because [2, 4] is a continuous subarray of size 2 and sums up to 6.

Example 2:

Input: [23, 2, 6, 4, 7],  k=6Output: TrueExplanation: 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.

题解：
从元素nums[i]到num[j],如果每个元素对k求余并累加，且累加后能被k整除，则TRUE，否则FALSE

具体代码如下：

class Solution {public:    bool checkSubarraySum(vector<int>& nums, int k) {        map<int, int> ma;        ma[0] = -1;        int q = 0;        for (int i = 0; i < nums.size(); ++i) {            q += nums[i];            if (k != 0)                q %= k;            if (ma.find(q) != ma.end()) {                if (i - ma[q] > 1)                    return true;            }            else                ma[q] = i;        }        return false;    }};

end!