Medium FB tag
naive方法：O(n2)

class Solution {
    public boolean checkSubarraySum(int[] nums, int k) {
        if (nums == null || nums.length == 0){
            return false;
        }            
        for (int i = 0; i < nums.length; i++){
            int sum = nums[i];
            for (int j = i + 1; j < nums.length; j++){
                sum += nums[j];
                if (k == 0){
                    return sum == 0;
                } else {
                    if (sum % k == 0){
                        return true;
                    }
                }
            }
        }
        return false;
    }
}

另一种是preSum的方法,不过这里要利用到余数的一些特性：
(a+(n*x))%x is same as (a%x),
就是如果preSum[6] % k = 1, 然后preSum[8] / k = 1, 又因为这个数列都是非负数，这种情况可以说明sum[7, 8] (inclusively)能被k整除。

class Solution {
    public boolean checkSubarraySum(int[] nums, int k) {
        if (nums == null || nums.length == 0){
            return false;
        }  
        // [0,0]
        // 0
        int preSum = 0;
        Map<Integer, Integer> preSumIndexMap = new HashMap<>();
        //[6,1,2,3] k=6 , when i = 0 won't return true coz prev = -1 and i = 0 so i - prev = 1 < 1
        preSumIndexMap.put(0, -1);
        for (int i = 0; i < nums.length; i++){
            preSum += nums[i];
            if (k != 0){
                preSum %= k;
            }
            if (preSumIndexMap.containsKey(preSum)){
                int prev = preSumIndexMap.get(preSum);
                if (i - prev > 1){
                    return true;
                }
            } else {
                preSumIndexMap.put(preSum, i);
            }
        }
        return false;
    }
}