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…

原题链接在这里:https://leetcode.com/problems/reverse-string-ii/#/description 题目: 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 l…

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…

344. Reverse String 最基础的旋转字符串 class Solution { public: void reverseString(vector<char>& s) { if(s.empty()) return; ; ; while(start < end){ char tmp = s[end]; s[end] = s[start]; s[start] = tmp; start++; end--; } return; } }; 541. Reverse Strin…

541. 反转字符串 II 541. Reverse String II…

Write a function that takes a string as input and returns the string reversed. Example: Given s = "hello", return "olleh". 这道题没什么难度,直接从两头往中间走,同时交换两边的字符即可,参见代码如下: 解法一: class Solution { public: string reverseString(string s) { , right =…

原题链接在这里:https://leetcode.com/problems/reverse-string/ 题目: Write a function that takes a string as input and returns the string reversed. Example:Given s = "hello", return "olleh". 题解: Reverse string是常见题, string在Java中是primitive类型, 一旦生成不…

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…

LeetCode--Reverse String Question Write a function that takes a string as input and returns the string reversed. Example: Given s = "hello", return "olleh". Answer class Solution { public: string reverseString(string s) { string str(s.…

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…