LeetCode - 523 - Continuous Subarray Sum

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:

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

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题意：给出一个数组和k值，求是否存在一个最长的子串和是k的倍数（子串长度大于1）

思路：唔，我实在不明白这题为啥划到dp里边了，想半天无从下手。网上大神给了一个o(n)的做法，统计前缀和%k，用map查询是否出现过这个值。出现过的话直接相减判断就好了，没出现过就丢进map里。服。

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