作者: 负雪明烛
id: fuxuemingzhu
个人博客: http://fuxuemingzhu.cn/


题目地址:https://leetcode.com/problems/reverse-words-in-a-string-iii/#/description

题目描述

Given a string, you need to reverse the order of characters in each word within a sentence while still preserving whitespace and initial word order.

Example:

Input: "Let's take LeetCode contest"
Output: "s'teL ekat edoCteeL tsetnoc"

Note:

In the string, each word is separated by single space and there will not be any extra space in the string.

题目大意

把字符串中的每个单词进行翻转,翻转后仍然按照原来的单词顺序进行拼接。

解题方法

Java解法

很简单的题,直接分割开每个单词,然后把单词翻转再拼接就好了。我的第一种做法:

public class Solution {
public String reverseWords(String s) {
String[] words = s.split(" ");
StringBuilder ans = new StringBuilder();
boolean isFirst = true;
for(String word : words){
StringBuilder temp = new StringBuilder(word);
word = temp.reverse().toString();
if(isFirst){
ans.append(word);
isFirst = false;
}else{
ans.append(" " + word);
} }
return ans.toString();
}
}

看着时间有点长 14 ms,于是没用StringBuilder,方法如下

public class Solution {
public String reverseWords(String s) {
String[] words = s.split(" ");
StringBuilder ans = new StringBuilder();
boolean isFirst = true;
for(String word : words){
word= reverse(word);
if(isFirst){
ans.append(word);
isFirst = false;
}else{
ans.append(" " + word);
} }
return ans.toString();
}
public String reverse(String s){
char[] chars = s.toCharArray();
for(int i = 0; i < chars.length / 2; i++){
char temp = chars[i];
chars[i] = chars[chars.length - 1 - i];
chars[chars.length - 1 - i] = temp;
}
return new String(chars);
}
}

时间变为 13 ms,还想继续压缩时间。全部用数组实现:

public class Solution {
public String reverseWords(String s) {
String[] words = s.split(" ");
char[] ans = new char[s.length()];
boolean isFirst = true;
int i = 0;
for (String word : words) {
char[] chars = word.toCharArray();
for (int j = 0; j < chars.length / 2; j++) {
char temp = chars[j];
chars[j] = chars[chars.length - 1 - j];
chars[chars.length - 1 - j] = temp;
}
System.arraycopy(chars, 0, ans, i, chars.length);
i += chars.length;
if (i != ans.length) {
ans[i] = ' ';
}
i++;
}
return new String(ans);
}
}

这个时间却变成了16ms,已经无语。嗯。就这样吧。

Python解法

使用Python可以直接使用split函数之后,进行[::-1]即做了翻转操作,然后再用" ".join()拼接在一起就行了。

class Solution(object):
def reverseWords(self, s):
"""
:type s: str
:rtype: str
"""
return " ".join(map(lambda x : x[::-1], s.split(" ")))

日期

2017 年 4 月 12 日
2018 年 11 月 6 日 —— 腰酸背痛要废了

【LeetCode】557. Reverse Words in a String III 解题报告(Java & Python)的更多相关文章

  1. LeetCode 557 Reverse Words in a String III 解题报告

    题目要求 Given a string, you need to reverse the order of characters in each word within a sentence whil ...

  2. Leetcode#557. Reverse Words in a String III(反转字符串中的单词 III)

    题目描述 给定一个字符串,你需要反转字符串中每个单词的字符顺序,同时仍保留空格和单词的初始顺序. 示例 1: 输入: "Let's take LeetCode contest" 输 ...

  3. leetcode 557. Reverse Words in a String III 、151. Reverse Words in a String

    557. Reverse Words in a String III 最简单的把空白之间的词反转 class Solution { public: string reverseWords(string ...

  4. [LeetCode] 557. Reverse Words in a String III 翻转字符串中的单词 III

    Given a string, you need to reverse the order of characters in each word within a sentence while sti ...

  5. Leetcode - 557. Reverse Words in a String III (C++) stringstream

    1. 题目:https://leetcode.com/problems/reverse-words-in-a-string-iii/discuss/ 反转字符串中的所有单词. 2. 思路: 这题主要是 ...

  6. LeetCode 557. Reverse Words in a String III (反转字符串中的单词 III)

    Given a string, you need to reverse the order of characters in each word within a sentence while sti ...

  7. 【leetcode】557. Reverse Words in a String III

    Algorithm [leetcode]557. Reverse Words in a String III https://leetcode.com/problems/reverse-words-i ...

  8. 557. Reverse Words in a String III【easy】

    557. Reverse Words in a String III[easy] Given a string, you need to reverse the order of characters ...

  9. Week4 - 500.Keyboard Row & 557.Reverse Words in a String III

    500.Keyboard Row & 557.Reverse Words in a String III 500.Keyboard Row Given a List of words, ret ...

随机推荐

  1. 通过mac地址确认二层交换机某个端口下接的终端设备IP

    正常来说,二层交换机主要是通过mac地址进行通信的,这就导致我们无法直接通过arp表来确认交换机端口下终端设备的IP: 但我们仍然能通过查找二层交换机端口下学习到的mac地址,然后通过对照三层核心交换 ...

  2. C4.5决策树-为什么可以选用信息增益来选特征

    要理解信息增益,首先要明白熵是什么,开始很不理解熵,其实本质来看熵是一个度量值,这个值的大小能够很好的解释一些问题. 从二分类问题来看,可以看到,信息熵越是小的,说明分类越是偏斜(明确),可以理解为信 ...

  3. SPI详解2

    串行外设接口 (SPI) 总线是一种运行于全双工模式下的同步串行数据链路.用于在单个主节点和一个或多个从节点之间交换数据. SPI 总线实施简单,仅使用四条数据信号线和控制信号线(请参见图 1). 图 ...

  4. flask分页功能:基于flask-sqlalchemy和jinja2

    先看源码: @app.route('/movie', methods=['GET', 'POST']) @app.route('/home', methods=['GET', 'POST']) @ap ...

  5. 日常Java 2021/9/28

    字符串反转 package m; public class m { public static void main(String[] args) { //定义一个字符串 String str = &q ...

  6. abide, able, abnormal

    abide 近/反义词:1. 忍受: bear, endure, put up with, stand, tolerate2. 遵守(abide by): accept, comply, confor ...

  7. 17. yum

    https://www.linuxidc.com/Linux/2015-04/116331.htm

  8. hadoop/spark面试题

    总结于网络 转自:https://www.cnblogs.com/jchubby/p/5449379.html 1.简答说一下hadoop的map-reduce编程模型 首先map task会从本地文 ...

  9. Hadoop的HA机制浅析

    Zookeeper在Hadoop的HA中的应用 非HA的弊端: HDFS集群的分布式存储是靠namenode节点(namenode负责响应客户端请求)来实现.在非HA集群中一旦namenode宕机,虽 ...

  10. 零基础学习java------day10------带包编译,权限修饰符,内部类,调式,junitTest

    0.  带包编译 解决使用notepad++编写的java类中如果有package的解决方案,如下代码 package com._51doit.test; class HelloWorld{ publ ...