LeetCode 523. Continuous Subarray Sum

原题链接在这里：https://leetcode.com/problems/continuous-subarray-sum/description/

题目：

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.

题解：

需要求有没有一段长度大于等于2的subarray的和是k的倍数.

保存之前每个点sum%k的值, 若是又出现了相同的值, 那么这一段subarray的和就是k的倍数.

然后看看这段长度是否大于等于2.

初始化HashMap<Integer, Integer> hm来记录到index i 时的sum%k 和 i的对应关系.

先赋值(0,-1). 原因是sum有包括nums[0]的情况.

这里通过初始index -1 和 i - hm.get(sum%k) > 1来确保. 

计算当前值的sum, 若k不是0, sum = sum%k. 看hm中是否已经有过这个余数, 若有看index差是否大于1, 没有就加进hm中.

答案有大于1的return true说明减掉之前的那段正好把余数减光. 若一直没有return false.

Note: check hm.containsKey(sum) and i - hm.get(sum) separately. Since later else should only applies for !hm.containsKey(sum).

Time Complexity: O(nums.length).

Space: O(nums.length).

AC Java:

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

类似Subarray Sum Equals K.