Leetcode 算法习题第十四周

最新推荐文章于 2022-07-29 18:25:44 发布

Cindy_bunny

最新推荐文章于 2022-07-29 18:25:44 发布

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本文链接：https://blog.youkuaiyun.com/Cindy_bunny/article/details/78738592

本文介绍了一种算法解决方案，用于判断一个非负整数数组中是否存在连续子数组，其元素之和为给定整数k的倍数。通过示例展示了如何使用该方法进行验证。

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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 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.

题目大意

判断数组中是否存在子数组使得加起来元素等于给定数的倍数

我的解答

class Solution {
public:
    bool checkSubarraySum(vector<int>& nums, int k) {
        int size = nums.size();
        vector<int> sum(nums.begin(),nums.end());//sum初始化为nums

        for(int i = 2; i <= size; i++){//从只有两个元素开始，到有size个元素相加
            for(int j = 0; j <= size-i; j++){//从第0个元素开始，到size-i个元素，取i个元素相加
                sum[j] += nums[j+i-1];//每次加一个
                if(sum[j] == 0)return true;//如果和为0，则为任何数的倍数
                else if(k != 0 && sum[j] % k == 0) return true;
            }
        }
        return false;

    }
};