LeetCode-Reverse Words in a String[AC源码]
package com.lw.leet1; import java.util.Stack; /**
* @ClassName:Solution
* @Description:
* Reverse Words in a String
* Total Accepted: 26194 Total Submissions: 187094 My Submissions
* Given an input string, reverse the string word by word.
*
* For example
* Given s = "the sky is blue"
* return "blue is sky the".
*
* Clarification:
* What constitutes a word?
* A sequence of non-space characters constitutes a word.
* Could the input string contain leading or trailing spaces?
* Yes. However, your reversed string should not contain leading or trailing spaces.
* How about multiple spaces between two words?
* Reduce them to a single space in the reversed string.
*
* @Author LiuWei
* @Date 2014年8月15日下午7:48:48
* @Mail nashiyue1314@163.com
*/
public class Solution { public String reverseWords(String word){
Stack<String> sstack = new Stack<String>();
int flag = 0;
for(int i= 0; i<word.length(); i++){
while(i<word.length() && word.charAt(i)==' '){
i++;
}
flag = i;
while(i<word.length() && word.charAt(i)!=' '){
i++;
}
if(flag != i){
sstack.push(word.substring(flag, i));
}
}
String res = "";
while(!sstack.isEmpty()){
res += sstack.pop()+" ";
}
// The input string which is made up of space
if(res.length()==0){
return "";
}
return res.substring(0, res.length()-1);
} public String reverseWords2(String word){
String res ="";
int flag = 0;
for(int i= 0; i<word.length(); i++){
while(i<word.length() && word.charAt(i)==' '){
i++;
}
flag = i;
while(i<word.length() && word.charAt(i)!=' '){
i++;
}
if(flag != i){
res = word.substring(flag, i)+" "+res;
}
}
// The input string which is made up of space
if(res.length()==0){
return "";
}
return res.substring(0, res.length()-1);
} public static void main(String[] args){
Solution s = new Solution();
System.out.println(s.reverseWords2(" hello world "));
}
}
LeetCode-Reverse Words in a String[AC源码]的更多相关文章
- LeetCode: Reverse Words in a String:Evaluate Reverse Polish Notation
LeetCode: Reverse Words in a String:Evaluate Reverse Polish Notation Evaluate the value of an arithm ...
- LeetCode Reverse Words in a String II
原题链接在这里:https://leetcode.com/problems/reverse-words-in-a-string-ii/ 题目: Given an input string, rever ...
- [LeetCode] Reverse Vowels of a String 翻转字符串中的元音字母
Write a function that takes a string as input and reverse only the vowels of a string. Example 1:Giv ...
- [LeetCode] Reverse Words in a String II 翻转字符串中的单词之二
Given an input string, reverse the string word by word. A word is defined as a sequence of non-space ...
- [LeetCode] Reverse Words in a String 翻转字符串中的单词
Given an input string, reverse the string word by word. For example, Given s = "the sky is blue ...
- [LeetCode] Reverse Words in a String III 翻转字符串中的单词之三
Given a string, you need to reverse the order of characters in each word within a sentence while sti ...
- LeetCode: Reverse Words in a String 解题报告
Reverse Words in a String Given an input string, reverse the string word by word. For example,Given ...
- java.lang.String 类源码解读
String类定义实现了java.io.Serializable, Comparable<String>, CharSequence 三个接口:并且为final修饰. public fin ...
- 翻String.Format源码发现的新东西:StringBuilderCache
起因: 记不清楚今天是为毛点想F12看String.Format的实现源码了,反正就看到了下图的鸟东西: 瞬间石化有没有,StringBuilder还能这么获取? 研究StringBuilderCac ...
随机推荐
- underscore.js源码解析(三)
最近工作比较忙,做不到每周两篇了,周末赶着写吧,上篇我针对一些方法进行了分析,今天继续. 没看过前两篇的可以猛戳这里: underscore.js源码解析(一) underscore.js源码解析(二 ...
- sql数据库表容量
标题:SQL Server 的最大容量规范 数据库的文件大小,文件数量都有限制. 表的大小也有限制,如果表过大,查询效率就会下降,考虑对数据进行分割,对历史数据进行独立存储.
- lintcode-439-线段树的构造 II
439-线段树的构造 II 线段树是一棵二叉树,他的每个节点包含了两个额外的属性start和end用于表示该节点所代表的区间.start和end都是整数,并按照如下的方式赋值: 根节点的 start ...
- lintcode-81-数据流中位数
81-数据流中位数 数字是不断进入数组的,在每次添加一个新的数进入数组的同时返回当前新数组的中位数. 说明 中位数的定义: 中位数是排序后数组的中间值,如果有数组中有n个数,则中位数为A[(n-1)/ ...
- Kotlint集合简单总结
1.数组操作 var testArray = Array<>("s","ss")或者 = arrayOf("s","s ...
- TP中循环遍历
循环遍历(重点) 在ThinkPHP中系统提供了2个标签来实现数组在模版中的遍历: volist标签.foreach标签. Volist语法格式: Foreach语法格式: 从上述的语法格式发现vol ...
- elasticsearch6 学习之基础CURD
环境:elasticsearch6.1.2 kibana6.1.2 基础概念: 1._index元数据 (1)代表一个document存放在哪个index中(2)类似的数据放在一个索引 ...
- json 当集合类型的字符串变成集合时候 里面有map类型的转换操作
- 集合里面的 E是泛型 暂且认为是object
集合里面的 E是泛型 暂且认为是object
- BZOJ 2462 矩阵模板(二维hash)
题意:给出一个n*m的01矩阵,以及k个a*b的01矩阵,问每个是否能匹配原来的01矩阵. 由于k个矩阵的长和宽都是一样的,所以把原矩阵的所有a*b的子矩阵给hash出来.然后依次查找是否存在即可. ...