用“%20”取代字符串中空格的时间复杂度为O(n)的算法

 1 /*length 为字符串数组string的总容量*/
 2 void ReplaceBlank(char stringp[],int length)
 3 {
 4   if(string==NULL &&length<=0)
 5       return;
 6 
 7   /*originalLength 为字符串string的实际长度*/
 8   int originalLength=0;
 9   int numberOfBlank=0;
10   int i=0;
11   while(string[i] !='')
12   {
13      ++originalLength;
14 
15      if(string[i]==' ')
16          ++numberOfBlank;
17      ++i;
18   }
19 
20   /*newlength 为把空格替换成%20之后的长度*/
21   int newLength=originalLength+numberOfBlank*2;
22   if(newLength >length)
23       return;
24 
25   int indexOfOriginal = originalLength;
26   int indexOfNe = newLength;
27   while(indexOfOriginal >=0 && indexOfNew >indexOfOriginal)
28   {
29   if(string[indexOfOriginal] ==' ')
30     {
31         string[indexOfNew --] ='0'; 
32         string[indexOfNew --] ='2'; 
33         string[indexOfNew --] ='%'; 
34     }
35     else
36     {
37        string[indexOfNew --]=string[indexOfOriginal];
38     }
39 
40       --indexOfOriginal;
41   }
42 }