整体上3个题都是求subarray,都是同一个思想,通过累加,然后判断和目标k值之间的关系,然后查看之前子数组的累加和. map的存储:560题是存储的当前的累加和与个数 561题是存储的当前累加和的余数与第一次出现这个余数的位置 325题存储的是当前累加和与第一次出现这个和的位置 其实561与325都是求的最长长度,那就一定要存储的是第一次出现满足要求的位置,中间可能还出现这种满足要求的情况,但都不能进行存储 560. Subarray Sum Equals K 求和为k的连续子数组的个数 h…

Given an array of integers and an integer k, you need to find the total number of continuous subarrays whose sum equals to k. Example 1: Input:nums = [1,1,1], k = 2 Output: 2 Note: The length of the array is in range [1, 20,000]. The range of numbers…

Given an array of integers and an integer k, you need to find the total number of continuous subarrays whose sum equals to k. Example 1: Input:nums = [1,1,1], k = 2 Output: 2 Note: The length of the array is in range [1, 20,000]. The range of numbers…

Given an array of integers and an integer k, you need to find the total number of continuous subarrays whose sum equals to k. Example 1: Input:nums = [1,1,1], k = 2 Output: 2 Note: The length of the array is in range [1, 20,000]. The range of numbers…

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 日期 题目地址:https://leetcode.com/problems/subarray-sum-equals-k/description/ 题目描述 Given an array of integers and an integer k, you need to find the total number of continuous subar…

Given an array of integers and an integer k, you need to find the total number of continuous subarrays whose sum equals to k. Example 1: Input:nums = [1,1,1], k = 2 Output: 2 Note: The length of the array is in range [1, 20,000]. The range of numbers…

［抄题］: Given an array of integers and an integer k, you need to find the total number of continuous subarrays whose sum equals to k. Example 1: Input:nums = [1,1,1], k = 2 Output: 2 [暴力解法]: 时间分析: 空间分析: [优化后]: 时间分析: 空间分析: ［奇葩输出条件］: ［奇葩corner case］: ［思维…

Given an array of integers and an integer k, you need to find the total number of continuous subarrays whose sum equals to k. Example 1: Input:nums = [1,1,1], k = 2 Output: 2 Note: The length of the array is in range [1, 20,000]. The range of numbers…

题目如下:解题思路:本题的关键在于题目限定了是连续的数组,我们用一个dp数组保存第i位到数组末位的和.例如nums = [1,1,1],那么dp = [3,2,1], dp[i]表示nums[i]+nums[i+1] +...+nums[len(nums)-1],有了这一个dp数组后,我们很容易就可以得到递推表达式 sum(i,j) = dp[i] - dp[j+1].最后,顺序遍历dp数组,对于任意的dp[i],只要找到对应的dp[k-i]就可以了. 代码如下: class Solution(…

Given an array nums and a target value k, find the maximum length of a subarray that sums to k. If there isn't one, return 0 instead. Example 1: Given nums = [1, -1, 5, -2, 3], k = 3,return 4. (because the subarray [1, -1, 5, -2] sums to 3 and is the…