从字符串中删除指定字符的有效方法

void Remove (char * src,const char * match) {
   char * dest = src;
   for (;;) { 
      char ch = *src++; 
      if (!strchr (match,ch)) *dest++ = ch;  // Copy chars that don't match
      if (!ch) break;                         // Stop when we copy over a null  
   }
}

我假设这些是空终止的.如果不是这种情况,那么您也必须传入长度并适当地修改算法.特别是,您将无法使用strchr.为了完整起见,这里有一个与char数组一起使用的版本(不是以null结尾).

// Removes from str[] (of length strlen),all chars that are found
// in match[] (of length matchlen). Modifies str in place,and returns
// the updated (shortened) length of str. 
int Remove (char[] str,int srclen,char[] match,int matchlen) {
   int dst = 0,found;
   for (int src = 0; src < srclen; src++) { 
      char ch = str[src];  
      found = 0;           // Search if this char is found in match
      for (int i = 0; i < matchlen && !found; i++) 
         if (match[i] == ch) found = 1;
      if (!found) str[dst++] = ch;
   }
   return dst;
}

最后,我认为这与我们将要获得的O(n)接近.我在这里假设8位字符并构建一个查找表,因此这应该在O(n)O(m)中运行,其中m是匹配字符串的长度.

int Remove (char* str,char* match,int matchlen) {
   bool found[256];
   for (int i = 0; i < 256; i++) found[i] = 0;
   for (int i = 0; i < matchlen; i++) found[match[i]] = 1; 

   int dst = 0;
   for (int src = 0; src < srclen; src++) { 
      char ch = str[src];  
      if (!found[ch]) str[dst++] = ch;
   }
   return dst;
}