leetcode523. Continuous Subarray Sum

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.

解法

同博客上一题leetcode560. sum[i]与sum[j]模k的余数相同，说明[i+1….j]的和为k的倍数。

public class Solution {
    public boolean checkSubarraySum(int[] nums, int k) {
        if (nums == null || nums.length == 0) {
            return false;
        }

        int sum = 0, result = 0;
        Map<Integer, Integer> preSum = new HashMap<>();
        preSum.put(0, -1);

        for (int i = 0; i < nums.length; i++) {
            sum += nums[i];
            if (k != 0) {
                sum %= k;
            }
            if (!preSum.containsKey(sum)) {
                preSum.put(sum, i);
            } else {
                if (i - preSum.get(sum) > 1) {
                    return true;
                }
            }
        }

        return false;
    }
}