LeetCode 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:

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.

题解：

DP问题. 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). 原因是加到 i 时可能sum%k是0, 为了确保subarray长度至少是2, 防止开头第一个值就是k的倍数情况.

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

状态转移, 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.

Time Complexity: O(nums.length). Space: O(nums.length).

AC Java:

1 class Solution {
2     public boolean checkSubarraySum(int[] nums, int k) {
3         HashMap<Integer, Integer> hm = new HashMap<Integer, Integer>();
4         hm.put(0,-1);
5
6         int sum = 0;
7         for(int i = 0; i<nums.length; i++){
8             sum += nums[i];
9             if(k != 0){
10                 sum = sum%k;
11             }
12
13             if(hm.containsKey(sum)){
14                 if(i-hm.get(sum)>1){
15                     return true;
16                 }
17             }else{
18                 hm.put(sum, i);
19             }
20         }
21         return false;
22     }
23 }

类似Subarray Sum Equals K.