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 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:
* The string consists of lower English letters only.
* Length of the given string and k will in the range [1, 10000]
*/
public class L541 {
public String reverseStr(String s, int k) {
StringBuilder sb = new StringBuilder();
int len = s.length();
int count = len/(2*k)-1;
int remaining = len%(2*k);
if (remaining!=0){
count = count+1;
}
for (int i=0;i<=count;i++){
if(remaining!=0 && i==count){
int cout2k = 2*k*(i+1);
int coutk = 2*k*i+k;
if(len<coutk){
StringBuilder sb3 = reverse(s,2*k*i,len);
sb.append(sb3);
}else if(len<cout2k){
StringBuilder sb3 = reverse(s,2*k*i,coutk);
sb.append(sb3);
StringBuilder sb4 = retain(s,coutk,len);
sb.append(sb4);
}
}else {
StringBuilder sb1 = reverse(s,2*k*i,2*k*i+k);
sb.append(sb1);
StringBuilder sb2 = retain(s,2*k*i+k,2*k*(i+1));
sb.append(sb2);
}
}
return new String(sb);
} private StringBuilder reverse(String s,int begin,int end){
return new StringBuilder(s.substring(begin,end)).reverse();
} private StringBuilder retain(String s,int begin,int end){
return new StringBuilder(s.substring(begin,end));
} }
class Solution {
public String reverseStr(String s, int k) {
char[] str=s.toCharArray();
int i;
int len=str.length;
for(i=0;i<len;i+=2*k){
if(i+k-1 >= len || i+(2*k)-1 >= len)
break;
reverse(str,i,i+k-1);
}
// if < 2k
if(i+k-1 < len)
reverse(str,i,i+k-1);
// if < k
else if(i< len && i+k-1 >= len)
reverse(str,i,len-1); return new String(str);
}
public void reverse(char[] str, int i, int j){
while(i<j){
char temp=str[i];
str[i]=str[j];
str[j]=temp;
i++;
j--;
} }
}
class Solution {
public String reverseStr(String s, int k) {
char[] a = s.toCharArray();
for (int start = 0; start < a.length; start += 2 * k) {
int i = start, j = Math.min(start + k - 1, a.length - 1);
while (i < j) {
char tmp = a[i];
a[i++] = a[j];
a[j--] = tmp;
}
}
return new String(a);
}
}

leadcode 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. 【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 ...

  4. 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 ...

  5. [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 ...

  6. 541. Reverse String II

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

  7. 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 ...

  8. 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 ...

  9. [LC] 541. Reverse String II

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

随机推荐

  1. echarts报表

    <%@ page language="java" contentType="text/html; charset=utf-8" pageEncoding= ...

  2. 确定文件的位置--浏览文件夹对话框folderBrowserDialog

    private void button1_Click(object sender, EventArgs e) { folderBrowserDialog1.ShowNewFolderButton = ...

  3. 你所不知道的JSON

    译者按: 老司机们,你知道JSON.stringify还有第二个和第三个可选参数吗?它们是什么呢? JSON已经逐渐替代XML被全世界的开发者广泛使用.本文深入讲解JavaScript中使用JSON. ...

  4. web服务器优化的一些思路

    作为一个新手(并不是菜鸟,而是像我们这样的学生),维护一个网站往往是一个很头疼的问题,尤其是动态网站,更尤其是用java写的网站. 当网站的吞吐量很小的时候你会发现服务器根本不需要维护,因为几乎没有延 ...

  5. JavaScript语言精粹读书笔记- JavaScript对象

    JavaScript 对象 除了数字.字符串.布尔值.null.undefined(都不可变)这5种简单类型,其他都是对象. JavaScript中的对象是可变的键控集合(keyed collecti ...

  6. Python之打印变量

    def myprint(v): print v print type(v) try: print v.shape except: try: print len(v) except: pass

  7. mac os x 记录 转载

    转载:远景网友(手机锋友t5sd3sf):http://bbs.feng.com/read-htm-tid-10434256.html 一个命令制作 OS X 原版安装U盘 1.要保证下载的原版安装包 ...

  8. 分页技巧__设计分页用的QueryHelper辅助对象

    分页技巧__设计分页用的QueryHelper辅助对象 FROM 必须 实体名 WHERE  可选 条件1 AND 条件2 AND ... 条件n ORDER BY 可选 属性1, 属性2, ... ...

  9. thinkphp5如何使用layout

    thinkphp5的layout和以前的版本有点差异. 首先开启配置文件中的 'template' => [      'layout_on' => true,      // 'layo ...

  10. 简单是Jedis实例(相对连接池而言)

    在引入相关jar包后,只要new一个Jedis对象,就能做redis相关操作了.以下是一个简单的jedis实例: package com.pptv.redis; import java.util.Ar ...