Replace the original spaces in the string

1: The theme of the program

1. Topic

　　Please implement a function to replace spaces in a string with "%20". For example, when the string is We Are Happy., the replaced string is We%20Are%20Happy.

2. My program

 1 package com.jianke.it;
 2 
 3 public class ReplaceSpace {
 4 
 5     public static void main(String[] args) {
 6         StringBuffer str=new StringBuffer("We Are Happy");
 7         String result=replaceSpace(str);
 8         System.out.println(result);
 9     }
10     public static String replaceSpace(StringBuffer str) {
11         StringBuffer str2=new StringBuffer();
12         int len=str.length();
13         for(int i=0;i<len;i++) {
14             if(str.charAt(i)!=' ') {
15                 str2.append(str.charAt(i));
16             }else {
17                 str2.append("%20");
18             }
19         }
20         
21         return str2.toString();
22     }
23 
24 }

3. Effects

4. Procedure 2

问题1：替换字符串，是在原来的字符串上做替换，还是新开辟一个字符串做替换！

问题2：在当前字符串替换，怎么替换才更有效率（不考虑java里现有的replace方法）。

从前往后替换，后面的字符要不断往后移动，要多次移动，所以效率低下

从后往前，先计算需要多少空间，然后从后往前移动，则每个字符只为移动一次，这样效率更高一点。

Conclusion: My method is to open up a new string, this method is to modify the original basis.

 1 package com.jianke.it;
 2 
 3 public class ReplaceSpace {
 4 
 5     public static void main(String[] args) {
 6         StringBuffer str=new StringBuffer("We Are Happy");
 7         String result=replaceSpace(str);
 8         System.out.println(result);
 9     }
10     public static String replaceSpace(StringBuffer str) {
11         //spacenum为计算空格数
12         int spacenum = 0;
13         for( int i=0;i<str.length();i++ ){
 14              if (str.charAt(i)==' ' )
 15                  spacenum++ ;
 16          }
 17          // indexold is the str subscript 
18          int before replacement indexold = str.length()-1 ; 
 19          // Calculate the str length 
20 after the space is converted into %20          int newlength = str.length() + spacenum*2 ;
 21          // indexold is after replacing the space with %20 The str subscript 
22          int indexnew = newlength-1 ;
 23          // Expand the length of str to the length after converting to %20 to prevent the subscript from going out of bounds 
24          str.setLength(newlength);
25         for(;indexold>=0 && indexold<newlength;--indexold){
26             if(str.charAt(indexold) == ' '){
27                 str.setCharAt(indexnew--, '0');
28                 str.setCharAt(indexnew--, '2');
29                 str.setCharAt(indexnew--, '%');
30             }else{
31                 str.setCharAt(indexnew--, str.charAt(indexold));
32             }
33         }
34         return str.toString();
35     }
36 
37}

5. Analysis

　　&& indexold<newlength

　　Mainly analyze this condition, if not, it will not be a problem, but why should it be added now.

　　I thought about it, it's like this:

　　If the following spaces have been added, and the indexold and indexnew are the same at this time, do not move any more, and the time complexity can be reduced.

Replace the original spaces in the string

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