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.

这种考subarray sum 常用到累加和数组啊, 要清楚从哪到哪开始求和, 在看题意怎么判断就ok了, 数组subarray sum 问题常常用累加和基础上改变

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

没想到用hashmap O(n) 即可, 关键是存的都是k的余数, 然后余数相见等于零即可 if (k != 0) runningSum %= k; 

public boolean checkSubarraySum(int[] nums, int k) {
    Map<Integer, Integer> map = new HashMap<Integer, Integer>();
    int runningSum = 0;
    for (int i=0;i<nums.length;i++) {
        runningSum += nums[i];
        if (k != 0) runningSum %= k; 
        if (runningSum == 0) {
            if (i > 0) {
                return true;
            }
        }
        Integer prev = map.get(runningSum);
        if (prev != null) {
            if (i - prev > 1) return true;
        }
        else map.put(runningSum, i);
    }
    return false;
}