［抄题］: Given a string and an integer k, you need to reverse the first k characters for every 2k characters counting from the start of the string. If there are less than k characters left, reverse all of them. If there are less than 2k but greater than…
problem 541. Reverse String II 题意: 给定一个字符串,每隔k个字符翻转这k个字符,剩余的小于k个则全部翻转,否则还是只翻转剩余的前k个字符. solution1: class Solution { public: string reverseStr(string s, int k) { int n = s.size(); int cnt = n / k; ; i<=cnt; i++) { ==) { if(i*k+k<n) reverse(s.begin()+i…
原题地址: 344 Reverse String: https://leetcode.com/problems/reverse-string/description/ 541 Reverse String II: https://leetcode.com/problems/reverse-string-ii/description/ 题目&&解法: 1.Reverse String: Write a function that takes a string as input and ret…
package leadcode; /** * 541. Reverse String II * Easy * 199 * 575 * * * Given a string and an integer k, you need to reverse the first k characters for every 2k characters counting from the start of the string. If there are less than k characters lef…
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 Java解法 Python解法 日期 题目地址:https://leetcode.com/problems/reverse-string-ii/#/description 题目描述 Given a string and an integer k, you need to reverse the first k characters for every…
给定一个字符串和一个整数 k,你需要对从字符串开头算起的每个 2k 个字符的前k个字符进行反转.如果剩余少于 k 个字符,则将剩余的所有全部反转.如果有小于 2k 但大于或等于 k 个字符,则反转前 k 个字符,并将剩余的字符保持原样.示例:输入: s = "abcdefg", k = 2输出: "bacdfeg"要求: 1.该字符串只包含小写的英文字母. 2.给定字符串的长度和 k 在[1, 10000]范围内.详见:https://leetcode.…
Given a string and an integer k, you need to reverse the first k characters for every 2k characters counting from the start of the string. If there are less than k characters left, reverse all of them. If there are less than 2k but greater than or eq…
Given a string and an integer k, you need to reverse the first k characters for every 2k characters counting from the start of the string. If there are less than k characters left, reverse all of them. If there are less than 2k but greater than or eq…
