用“%20”取代字符串中空格的时间复杂度为O(n)的算法

 /*length 为字符串数组string的总容量*/
 void ReplaceBlank(char stringp[],int length)
 {
   if(string==NULL &&length<=)
       return;

   /*originalLength 为字符串string的实际长度*/
   int originalLength=;
   int numberOfBlank=;
   int i=;
   while(string[i] !='\0')
   {
      ++originalLength;

      if(string[i]==' ')
          ++numberOfBlank;
      ++i;
   }

   /*newlength 为把空格替换成%20之后的长度*/
   int newLength=originalLength+numberOfBlank*;
   if(newLength >length)
       return;

   int indexOfOriginal = originalLength;
   int indexOfNe = newLength;
   while(indexOfOriginal >= && indexOfNew >indexOfOriginal)
   {
   if(string[indexOfOriginal] ==' ')
     {
         string[indexOfNew --] ='';
         string[indexOfNew --] ='';
         string[indexOfNew --] ='%';
     }
     else
     {
        string[indexOfNew --]=string[indexOfOriginal];
     }

       --indexOfOriginal;
   }
 }