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.

preSum一下，然后做差判断是否符合条件

代码：

    public boolean checkSubarraySum(int[] nums, int k) {
        if(nums == null || nums.length < 2) return false;
        
        int[] preSum = new int[nums.length+1];
        int sum = 0;
        for(int i=1;i<=nums.length;i++) {
            sum += nums[i-1];
            preSum[i] = sum;
        }
        
        for(int i=0;i<nums.length;i++) {
            for(int j=i+2;j<=nums.length;j++) {
                if(k == 0 && ((preSum[j] == preSum[i]))) return true;  
                if(k != 0 && (((preSum[j] - preSum[i]) % k) == 0)) return true;
            }
        }
        return false;
    }