523. Continuous Subarray Sum

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Problem

Given an integer array nums and an integer k, return true if nums has a continuous subarray of size at least two whose elements sum up to a multiple of k, or false otherwise.

An integer x is a multiple of k if there exists an integer n such that x = n * k. 0 is always a multiple of k.

Example 1:

Input: nums = [23,2,4,6,7], k = 6
Output: true
Explanation: [2, 4] is a continuous subarray of size 2 whose elements sum up to 6.

Example 2:

Input: nums = [23,2,6,4,7], k = 6
Output: true
Explanation: [23, 2, 6, 4, 7] is an continuous subarray of size 5 whose elements sum up to 42.
42 is a multiple of 6 because 42 = 7 * 6 and 7 is an integer.

Example 3:

Input: nums = [23,2,6,4,7], k = 13
Output: false

Constraints:

• 1 <= nums.length <= 10^5
• 0 <= nums[i] <= 10^9
• 0 <= sum(nums[i]) <= 2^31 - 1
• 1 <= k <= 2^31 - 1

Code

560. Subarray Sum Equals K

class Solution {
    public boolean checkSubarraySum(int[] nums, int k) {
        HashMap<Integer, Integer> map = new HashMap<>();
        
        int sum = nums[0] % k;
        map.put(sum, 0);
        
        for(int i = 1; i < nums.length; i++) {
            int num = nums[i];
    
            sum = (sum + num) % k;
            
            // [23,2,4,6,6], 7 -> [2,4,1,0,6]
            if(sum == 0) return true;
            if(map.containsKey(sum) && i - map.get(sum) > 1) return true;
            
            // [1,3,0,6], 6
            if(!map.containsKey(sum)) {
               map.put(sum, i); 
            }
            
        }
        
        return false;
    }
}