把字符串中的空格替换为"%20"

这个需要注意的是字符串的结尾最后一个字符为'\0'，并不是空字符，复制时要一块复制，算法思想就是先计算出字符串中总的空格数，然后

重新计算字符串的长度，由于"%20"为3个字符，比原来多2个，所以，字符串长度是原来字符串长度加上空格字符总数×2,就是新的字符串的长度。

代码如下:

#include <iostream>
#include <cstdlib>
using namespace std;
 
void  strReplace(char str[],int length)
  {
          if(str==NULL||length<0)
             exit(-1);
           int blank=0;
           int nLength;
          for(int i=0;i<length;i++)
             {
                  
             if(str[i]==' ')
                blank++;
             }
            nLength=length+blank*2;
          for(int j=length;j>=0;j--)
             {
                 if(str[j]!=' ')
                   {
                      str[nLength]=str[j];
                        nLength--; 
                   }
                  else
                   {
                       str[nLength--]='0';
                       str[nLength--]='2';
                       str[nLength--]='%';
                   }
                
                      
             }
               
            
  }
 int main()
{
     char str[13]="we are happy";
     int length=13;
     cout<<"before replace:  ";
     cout<<str<<endl;
     strReplace(str,13);
     cout<<"after replace:  ";
     cout<<str<<endl;
     return 0;
      
}