[LeetCode] 523. Continuous Subarray Sum

stackfing 发布于2019-08-16 13:46 / 790人阅读

Problem

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

class Solution {
    public boolean checkSubarraySum(int[] nums, int k) {
        if (nums == null || nums.length < 2) return false;
        if (k == 0) {
            for (int i = 0; i < nums.length-1; i++) {
               if (nums[i] == 0 && nums[i+1] == 0) return true;
            }
            return false;
        }
        Map map = new HashMap<>();
        int sum = 0;
        map.put(0, -1);
        for (int i = 0; i < nums.length; i++) {
            sum += nums[i];
            sum %= k;
            if (map.containsKey(sum)) {
                if (i-map.get(sum) > 1) return true;
            } else {
                map.put(sum, i);
            }
        }
        return false;
    }
}

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