删除所有出现在C字符串中的字符-需要示例

Question

InputString: "I am unwell" "We need to go to the doctor" "How long will it take?" .

OutputString: I am unwell We need to go to the doctor How long will it take?

The string needs to cleaned of all occurrences of the char " . I can think of the following approacg

1. Use, strchr() function finding first occurrence of "
2. Move all characters in the string left by once position.

Repeat steps 1 and 2 , until strchr() returns a NULL pointer.

I feel this is very inefficient way to approach this problem. I need to know , if there are other methods to achieve this? Pseudo code or actual code will both be appreciated.

Answer 1

for (s=d=str;*d=*s;d+=(*s++!='"'));

Answer 2

You can accomplish this by visiting each char of the string once. You basically copy the string over itself, skipping the " characters:

pseudocode:

1. Start with two pointers: SOURCE and DESTINATION. They both point to the first char of the string.
2. If *SOURCE == NULL set *DESTINATION = NULL. Stop.
3. If *SOURCE != " set *DESTINATION = *SOURCE and increment DESTINATION.
4. Increment SOURCE. Go to step 2.

code:

// assume input is a char* with "I am unwell\" \"We need to go..."

char *src, *dest;

src = dest = input;    // both pointers point to the first char of input
while(*src != '\0')    // exit loop when null terminator reached
{
    if (*src != '\"')  // if source is not a " char
    {
        *dest = *src;  // copy the char at source to destination
        dest++;        // increment destination pointer
    }
    src++;             // increment source pointer
}
*dest = '\0';          // terminate string with null terminator              

// input now contains "I am unwell We need to go..."

update : fixed some bugs in code

Answer 3

Instead of moving the characters "in-place" to overwrite the character being deleted, create a new string.

This minimizes the number of characters copied by copying each valid character once. With the original method, characters near the end of the string are copyied n times, where n is the number of invalid characters preceding it.

Answer 4

If your string is not very big, the obvious answer would be to have a separate string. A single loop till you get \\0 (End of string) Have a loop (Gives you O(n)) and a comparison to check if the current location of the string is the character in question (again O(n) )

In all :

---

  s1 = original array
  s2 = new array to store the final result
  c = character in question.  
  current_pointer = 0 
  new_pointer =0 
  while(s1[current_pointer] != '\0') {
   ele = s1[current_pointer] ;

   if( ele != c)  { 
    s2[new_pointer++] = ele
   }
    current_pointer++
  }

---

Note that this method works only when string sizes are small. We need to go for better methods as size of string increases.

Hope this helps.