[刷题防痴呆] 0523 - 连续的子数组和 (Continuo

题目地址

https://leetcode.com/problems/continuous-subarray-sum/

题目描述

523. Continuous Subarray Sum

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

思路

• 前缀和.
• 同余定理： subNum % k == 0，等价于 sum[j+1] % k == sum[i] % k.

关键点

代码

• 语言支持：Java

class Solution {
    public boolean checkSubarraySum(int[] nums, int k) {
        int n = nums.length;
        int[] sums = new int[n + 1];
        for (int i = 1; i <= n; i++) {
            sums[i] = sums[i - 1] + nums[i - 1];
        }
        Map<Integer, Integer> map = new HashMap<>();
        for (int i = 0; i <= n; i++) {
            int sumMod = sums[i] % k;
            if (map.containsKey(sumMod) && i > map.get(sumMod) + 1) {
                return true;
            }
            map.putIfAbsent(sumMod, i);
        }

        return false;
    }
}