[LeetCode] Problem 523 - Continuous Subarray Sum

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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

No.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.

No.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.

Code

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// 若数字a和b分别除以数字c，若得到的余数相同，那么(a-b)必定能够整除c
public boolean checkSubarraySum(int[] nums, int k) {
    int sum = 0;
    Map<Integer, Integer> map = new HashMap<>();
    map.put(0, -1);

    for (int i = 0; i < nums.length; i++) {
        sum += nums[i];
        int mod = (k == 0) ? sum : sum % k;

        if (map.containsKey(mod) && i - map.get(mod) > 1)
            return true;
        else if (!map.containsKey(mod))
            map.put(mod, i);
    }

    return false;
}