在字符串S1中删除字符串S2中所包含的字符

 /*************************************************************************
     > File Name: test.c
     > Author: ToLiMit
     > Mail: [e-mail]
     > Created Time: Sun 04 Jan 2015 06:20:05 PM PST
  ************************************************************************/

 #include<stdio.h>

 void delete_str_char (char * main_str, char * sub_str)
 {
     if ((main_str == NULL) || (sub_str == NULL))
         return;

     char * sub_index = sub_str;
     char * main_index = main_str;
     char bitmap[] = {};
     char * str = (char *)malloc (strlen (main_str) + );
     char * index = str;
     memset (str, , strlen (main_str) + );

     while (*sub_index != '\0') {
         char suffix = ((*sub_index) / ) - ;
         char offset = (*sub_index) % ;

         bitmap[suffix] |= (0x1 << ( - offset));
         sub_index++;
     }

     while (*main_index != '\0') {
         char suffix = ((*main_index) / ) - ;
         char offset = (*main_index) % ;

         if ((bitmap[suffix] & (0x1 << ( - offset))) == ) {
             *index = *main_index;
             index++;
         }
         main_index++;
     }

     *index = '\0';
     memcpy (main_str, str, strlen (str) + );
     free (str);
     return;
 }

 int main (int argc, char * argv[])
 {
     char test[] = "aabcdaaaaabcaacb";

     delete_str_char (test, "bcd");
     printf ("%s\n", test);
     return ;
 }

1. 用bitmap标记sub_str中出现过的字符

2. 申请一段buffer, 长度与main_str一致

3. 遍历main_str, 如果字符没有出现在bitmap中, 就将字符写入到buffer中

4. 遍历结束后, buffer中的数据就是删掉了sub_str中出现过的数据

时间复杂度: O(n + m) n为main_str长度, m为sub_str长度

空间复杂度: O(n)