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.

Solution：

思路：
We iterate through the input array exactly once, keeping track of the running sum mod k of the elements in the process. If we find that a running sum value at index j has been previously seen before in some earlier index i in the array, then we know that the sub-array (i,j] contains a desired sum.

Time Complexity: O(N) Space Complexity: O(N)

Solution Code：

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