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 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
A sorted list A contains 1, plus some number of primes. Then, for every p < q in the list, we consider the fraction p/q. What is the K-th smallest fraction considered? Return your answer as an array of ints, where answer[0] = pand answer[1] = q. Ex
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 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 th
题目连接 题意:在大小为1e5以内的数组求存在多少个区间和的值等于k的次方 这种题很经常见,总是想着用两个for循环解决,但是一定会超时. 题解:算出前缀和,使用map去查找mp[sum[i+1]-tmp]的个数,加起来就是答案,这样复杂度在O(n)加上mp的查找时间,基本上不会超时 #include<bits/stdc++.h> using namespace std; typedef long long ll; typedef long double db; typedef pair<
Given a linked list, reverse the nodes of a linked list k at a time and return its modified list. k is a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple of k then left-out nodes in
题目 输入 n 个整数,找出其中最小的 k 个数.例如输入4.5.1.6.2.7.3.8 这8个数字,则最小的4个数字是1.2.3.4. 初窥 这道题最简单的思路莫过于把输入的 n 个整数排序,排序之后位于最前面的 k 个数就是最小的 k 个数.这种思路的时间复杂度是 O(nlogn). 解法一:脱胎于快排的O(n)的算法 如果基于数组的第 k 个数字来调整,使得比第 k 个数字小的所有数字都位于数组的左边,比第 k 个数字大的所有数字都位于数组的右边.这样调整之后,位于数组中左边的 k 个数字