java 按字母顺序排列给定字符串中的单词
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/15597223/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Arranging words within a given string alphabetically
提问by Ashin Banerjee
i'm just trying to write a code for arranging the words within a string alphabetically.. But whenever i am running this it' entering in an infinite loop.. i cant figure out what is exactly happening..can anybody help me out plzzz.. below i've attached my code.
我只是想写一个代码来按字母顺序排列字符串中的单词..但是每当我运行它时，它就会进入一个无限循环..我无法弄清楚到底发生了什么..有人能帮我解决吗？ .. 下面我附上了我的代码。
public class AscendString {
String s=new String();
public AscendString(String x)
{
s=x.trim();
}
public int NoWords()
{
int i=0;
String s1=new String();
s1=s;
while(s1.length() > 0)
{ i++;
int j=s1.indexOf(' ');
if(j>0)
{
s1.substring(j+1);
s1=s1.trim();
}
else
s1="";
}
return i;
}
public void Ascend()
{
String str[]=new String[NoWords()];
String s1=new String();
s1=s;
int i=0;
while(s1.length() > 0)
{
int j=s1.indexOf(' ');
if(j>0)
{
str[i]=s1.substring(0,j) ;
s1=s1.substring(j+1);
s1=s1.trim();
i++;
}
else
{
str[i]=s1;
s1="";
}
}
for(int j=0;j < str.length-1;j++)
{
for(int k=0;k < str.length-1-j;k++)
if(str[k].length() > str[k+1].length())
{String temp=str[k];
str[k]=str[k+1];
str[k+1]=temp;
}
}
String str1="";
for(int n=0;n < str.length;n++)
str1=str1+str[n] +" " ;
System.out.println("The String in Alphabetic Order is: "+str1);
}
public static void main(String args[])
{
AscendString exmpl=new AscendString("I Love Java Programming");
exmpl.Ascend();
}
}
回答by Cratylus
Why are you making this hard?
你为什么要这么难？
String exmpl=new String("I Love Java Programming");
String[] parts = exmpl.split("\s+");
Arrays.sort(parts);
StringBuilder sb = new StringBuilder();
for(String s:parts){
sb.append(s);
sb.append(" ");
}
String sorted = sb.toString.trim();
回答by BobTheBuilder
Try:
尝试：
s1 = s1.substring(j+1);
substringreturn the sub-string, not changing the string., so this loop:
substring返回子字符串，不改变字符串。，所以这个循环：
while(s1.length() > 0)
{ i++;
int j=s1.indexOf(' ');
if(j>0)
{
s1.substring(j+1);
s1=s1.trim();
}
else
s1="";
}
Goes forever.
永远去。
回答by Eric Jablow
- There is never any reason to call
new String(...). - Should punctuation matter?
- What about repeated words?
- Do you want strings to sort caring about case? What do you expect from "A man, a plan, a canal, Panama."? "A Panama. a a canal man, plan,"
- 从来没有任何理由打电话
new String(...)。 - 标点符号重要吗？
- 重复的单词怎么办？
- 你想要字符串排序关心大小写吗？你对“一个人，一个计划，一条运河，巴拿马”有什么期望？“一个巴拿马。一个运河人，计划，”
Adapting Cratylus' answer, you might change the regex to \\W+. You'd be splitting on runs of non-word-constituents. And then, learn about Collatorand CollationKey. You probably want case-insensitive searches. Or, lower-case the string on the way in.
根据 Cratylus 的回答，您可能会将正则表达式更改为\\W+. 你会分裂非单词成分的运行。然后，了解Collator和CollationKey。您可能想要不区分大小写的搜索。或者，将输入的字符串小写。
