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.

第一种解法，暴力枚举所有可能子数组和，并判断，时间复杂度O(n^2)

class Solution {
public:
    bool checkSubarraySum(vector<int>& nums, int k) {
    if(k == 0){
        for(int i = 0;i < nums.size() - 1;i++)
            if(nums[i] == 0 && nums[i + 1] == 0)
                return true;
        return false;
    }
    int sum[nums.size()];
    for(int i = 0; i < nums.size(); i++){
        sum[i] = nums[i];
        for(int j = i + 1;j < nums.size(); j++){
            sum[i] += nums[j];
            if(sum[i] % k == 0)
                return true;
        }
    }
    return false;
}
};

第二种解法，使用一个unordered map记录模k的值的下标，当发现有k值相同的下标值，判断两个下标距离是否大于1，若距离大于1，则判断为真，unordered map查找元素时间复杂度为O(1)，总的时间复杂度为O(n)

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