# LeetCode Continuous Subarray Sum

LeetCode 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:

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.

• [23,25,29,35,42]
• [5,1,5,5,0]

```class Solution {
public:
bool checkSubarraySum(vector<int>& nums, int k) {
unordered_map<int, int> reminders;
reminders[0] = -1;
int accusum = 0;
for (int i = 0; i < nums.size(); ++i) {
accusum += nums[i];
if (k != 0)accusum %= k; // 注意k为0时，不能取模
if (reminders.find(accusum) != reminders.end()) {
if (i - reminders[accusum] >= 2)return true;
} else reminders[accusum] = i; // 注意必须是在else分支
}
return false;
}
};
```