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 equal to k characters, then reverse the first k characters and left the other as original.

Example:

Input: s = "abcdefg", k = 2
Output: "bacdfeg"

Restrictions:

  1. The string consists of lower English letters only.
  2. Length of the given string and k will in the range [1, 10000]
class Solution {
public String reverseStr(String s, int k) {
int i = 0;
char[] charArr = s.toCharArray();
while (i < charArr.length) {
int j = Math.min(i + k - 1, charArr.length - 1);
swap(i, j, charArr);
i += 2 * k;
}
return new String(charArr);
} private void swap(int i, int j, char[] charArr) {
while (i < j) {
char temp = charArr[i];
charArr[i] = charArr[j];
charArr[j] = temp;
i += 1;
j -= 1;
}
}
}

[LC] 541. Reverse String II的更多相关文章

  1. [LeetCode] 344 Reverse String && 541 Reverse String II

    原题地址: 344 Reverse String: https://leetcode.com/problems/reverse-string/description/ 541 Reverse Stri ...

  2. leetcode 344. Reverse String 、541. Reverse String II 、796. Rotate String

    344. Reverse String 最基础的旋转字符串 class Solution { public: void reverseString(vector<char>& s) ...

  3. leadcode 541. Reverse String II

    package leadcode; /** * 541. Reverse String II * Easy * 199 * 575 * * * Given a string and an intege ...

  4. 【leetcode_easy】541. Reverse String II

    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: cl ...

  5. LeetCode 541. Reverse String II (反转字符串 II)

    Given a string and an integer k, you need to reverse the first k characters for every 2k characters ...

  6. 541. Reverse String II 指定翻转前k个的字符串

    [抄题]: Given a string and an integer k, you need to reverse the first k characters for every 2k chara ...

  7. [LeetCode&Python] Problem 541. Reverse String II

    Given a string and an integer k, you need to reverse the first k characters for every 2k characters ...

  8. 541. Reverse String II

    static int wing=[]() { std::ios::sync_with_stdio(false); cin.tie(NULL); ; }(); class Solution { publ ...

  9. 541 Reverse String II 反转字符串 II

    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 ...

随机推荐

  1. 干货|CVE-2019-11043: PHP-FPM在Nginx特定配置下任意代码执行漏洞分析

    近期,国外安全研究员Andrew Danau,在参加夺旗赛(CTF: Capture the Flag)期间,偶然发现php-fpm组件处理特定请求时存在缺陷:在特定Nginx配置下,特定构造的请求会 ...

  2. py02_04:三元运算法

    a if a > b else c     #  a>b 成立,则为真,如果a>b为假,则返回c

  3. keras_yolo3阅读

    源码地址 https://github.com/qqwweee/keras-yolo3 春节期间仔细看了看yolov3的kears源码,这个源码毕竟不是作者写的,有点寒酸,可能大道至简也是这么个理.我 ...

  4. 静态页面缓存(thymeleaf模板writer)

    //前端html <!DOCTYPE html><html lang="en"> <head> <meta charset="U ...

  5. 三、VIP课程:并发编程专题->01-并发编程之Executor线程池详解

    01-并发编程之Executor线程池详解 线程:什么是线程&多线程 线程:线程是进程的一个实体,是 CPU 调度和分派的基本单位,它是比进程更小的能独立运行的基本单位.线程自己基本上不拥有系 ...

  6. Bless All

    # php code $i = 2333 $myJXOI = JXOI() while($i == 2333){ ++myJXOI.score , ++myJXOI.rp , --myJXOI.常数 ...

  7. typescript--介绍ts

    TypeScript 介绍 TypeScript 是什么 TypeScript 是 JavaScript 的强类型版本.然后在编译期去掉类型和特有语法,生成纯粹的 JavaScript 代码.由于最终 ...

  8. C#匿名委托,匿名函数,lambda表达式

    一.类型.变量.实例之间的关系. 类型>变量>实例 类型可以创建变量,实体类可以创建实例,实例可以存储在变量里. 二.委托使用过程: 1.定义委托(写好签名): 2.创建委托变量: 3.给 ...

  9. selenium自学记录2014.12.26

    现在还是有点病急乱投医的感觉,不知道到底该从何学起,到底怎么学 手头上有的资料是 <零成本实现Web自动化测试-基于Selenium和Bromine> <Selenium测试实践-基 ...

  10. TPO1-3Timberline Vegetation on Mountains

    At the upper timberline the trees begin to become twisted and deformed. This is particularly true fo ...