LeetCode #523 - Continuous Subarray Sum

题目描述：

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.

子数组之和要等于n*k，那么只需要用哈希表保存前缀和模k的结果即可。

class Solution {
public:
    bool checkSubarraySum(vector<int>& nums, int k) {
        if(nums.size()==0) return false;
        unordered_map<int,int> hash;
        hash[0]=-1; // 必须加上hash[0]=-1，表示前缀和从0开始
        int sum=0;
        for(int i=0;i<nums.size();i++)
        {
            sum+=nums[i];
            int temp=0;
            if(k==0) temp=sum; // 当k等于0时，不能求模
            else temp=sum%k;
            if(hash.count(temp)==0) hash[temp]=i;
            else if(i-hash[temp]>1) return true;
        }
        return false;
    }
};