将字符串中的空格替换为%20

分析：若从前往后遍历字符串插入%20移动的位数较多，每经过一次空格插入后，空格后面的字符都会往后移动

因此尝试从字符串后面往前面遍历，这样事先计算出需要的字符位数，再将%20插入，这样移动的位数较少

public class solution {

public static void main(String[] args) {

StringBuffer str1 = new StringBuffer("we are family");

// String str1 = "we are family";
System.out.println(replaceSpace(str1));

}

public static String replaceSpace(StringBuffer str){
int i = 0;
int oldnumber = 0;
int spacenumber = 0;
char[] string = str.toString().toCharArray();

// char[] string = str.toCharArray();

String newstring;

while(i<string.length){
if(string[i]==' '){
spacenumber++;
}
oldnumber++;
i++;
}
int newnumber = oldnumber+spacenumber*2;
char newstr[] = new char[newnumber];
int Poldnumber = oldnumber-1;
int Pnewnumber = newnumber-1;
while((Pnewnumber>=Poldnumber)&&(Pnewnumber>=0)){
if(string[Poldnumber]!=' '){
newstr[Pnewnumber--] = string[Poldnumber];
}
else {
newstr[Pnewnumber--] = '0';
newstr[Pnewnumber--] = '2';
newstr[Pnewnumber--] = '%';
}
Poldnumber--;
}
newstring = String.valueOf(newstr);
return newstring;
}

}