# 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 a 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.

DP(比较奇怪)

Corner Case:
k == 0, 0不能做除数

Pending:
HashMap算法

O(N2) O(N)

## Code

``````class Solution {
public boolean checkSubarraySum(int[] nums, int k) {
if(nums == null || nums.length == 0)
return false;

int[] dp = new int[nums.length];
dp = nums;
for(int i = 1; i < nums.length; i++){
dp[i] = dp[i-1] + nums[i];
}

for(int i = 0; i < nums.length; i++){
for(int j = i+1; j<nums.length; j++){
int diff = i - 1 >= 0 ? dp[j] - dp[i-1] : dp[j];
if(k == 0){
if(diff == 0)
return true;
}else {
if (diff % k == 0)
return true;
}
}
}
return false;
}
}
``````