LeetCode 523 Continuous Subarray Sum (同余)

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.

同余的两段相减后的段和模k必为0，注意特判k等于0的情况，击败82%

public class Solution {
    public boolean checkSubarraySum(int[] nums, int k) {
        int n = nums.length;
        if (n < 2) {
            return false;
        }
        if (k == 0) {
            for (int i = 0; i < n - 1; i ++) {
                if (nums[i] == 0 && nums[i + 1] == 0) {
                    return true;
                }
            }
            return false;
        }
        if (k == 1) {
            return true;
        }
        Set st = new HashSet<>();
        int sum = nums[0];
        st.add(sum % k);
        for (int i = 1; i < n; i ++) {
            sum += nums[i];
            sum %= k;
            if (sum == 0 || st.contains(sum)) {
                return true;
            }
            st.add(sum);
        }
        return false;
    }
}