[LeetCode] 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 a 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.

Constraints:

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.

连续的子数组和。题意是给定一个包含非负数的数组和一个目标整数 k，编写一个函数来判断该数组是否含有连续的子数组，其大小至少为 2，总和为 k 的倍数，即总和为 n*k，其中 n 也是一个整数。

思路是前缀和。注意题目的要求，子数组的长度至少为2，所以类似这样的情形，应该是要return false的。

input: [0], k = 0

既然是前缀和的思路，所以还是需要创建一个hashmap，存前缀和以及对应的index。既然是找sum为K的倍数，所以自然而然想到取模的操作。每次发现有一个前缀和 % K == 0的时候，去判断这个前缀和 % K是否存在于hashmap，若存在，再比较当前index和map.value的差值是否大于2，若是则return true。遍历结束则return false。

时间O(n)

空间O(n)

Java实现

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

JavaScript实现

 1 /**
 2  * @param {number[]} nums
 3  * @param {number} k
 4  * @return {boolean}
 5  */
 6 var checkSubarraySum = function(nums, k) {
 7     let sum = 0;
 8     let map = new Map();
 9     map.set(0, -1);
10     for (let i = 0; i < nums.length; i++) {
11         sum += nums[i];
12         if (k !== 0) {
13             sum %= k;
14         }
15         let prev = map.get(sum);
16         if (prev !== undefined) {
17             if (i - prev > 1) {
18                 return true;
19             }
20         } else {
21             map.set(sum, i);
22         }
23     }
24     return false;
25 };

前缀和prefix sum题目总结

LeetCode 题目总结