[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

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

JavaScript实现

/**
 * @param {number[]} nums
 * @param {number} k
 * @return {boolean}
 */
var checkSubarraySum = function(nums, k) {
    let sum = 0;
    let map = new Map();
    map.set(0, -1);
    for (let i = 0; i < nums.length; i++) {
        sum += nums[i];
        if (k !== 0) {
            sum %= k;
        }
        let prev = map.get(sum);
        if (prev !== undefined) {
            if (i - prev > 1) {
                return true;
            }
        } else {
            map.set(sum, i);
        }
    }
    return false;
};

前缀和prefix sum题目总结

LeetCode 题目总结