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.

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

两个数如果相差k的整数倍，那么这两个数相对于k的余数是一样的，因此，将sum对于k取余即可。。。