检查字符串是否在C中是回文 [英] Check if a string is palindrome in C

问题描述

我对我正在练习的这段代码有疑问.我必须检查一个字符串是否是回文.我无法更改该函数的声明.只有当所有字母都相同时该函数才返回1(例如"aaaa")，但是如果我将该句子归类为其他回文(例如"anna")，则该函数返回0，我不知道为什么会这样.谢谢！

i've a question about this code i'm writing for an exercise. I've to check if a string is palindrome. I can't change the declaration of the function.The function only return 1 when all the letters are the same (like "aaaa") but if i charge the sentence with other palindrome (like "anna") the function return me 0 , i can't figure out why this appening.Thank you!

char* cargar (char*);
int pali (char*);

int main()
{ 
   char*texto=NULL;
   texto=cargar(texto);
   int res=pali(texto);
   if(res==1){printf("\nPalindrome");}
   else printf("\nNot palindrome");

   return 0;
}

char* cargar (char*texto)
{
   char letra;
   int i=0;
   texto=malloc(sizeof(char));
   letra=getche();
   *(texto+i)=letra;
   while(letra!='\r'){
      i++;
      texto=realloc(texto,(i+1)*sizeof(char));
      letra=getche();
      *(texto+i)=letra;}
   *(texto+i)='\0';      
   return texto;
}

int pali (char* texto)
{
   int i;
   for(i=0;*(texto+i)!='\0';i++){
   }i--;
   if(i==0||i==1){return 1;}

   if(*texto==*(texto+i)){
      return pali(++texto);
   }
   else return 0;
}

推荐答案

确定字符串是否为回文式的函数的想法不完善.

Your function to determine whether a string is a palindrome is not well thought out.

假设您的字符串s的长度为l.字符串中的字符布局为:

Let's say you have a string s of length l. The characters in the string are laid out as:

Indices: 0    1    2    3            l-4  l-3  l-2  l-1
         +----+----+----+----+- ... -+----+----+----+----+
         |    |    |    |    |  ...  |    |    |    |    |   
         +----+----+----+----+- ... -+----+----+----+----+

如果字符串是回文，

s[0] = s[l-1]
s[1] = s[l-2]

...

当LHS的索引大于或等于LHS时，您可以停止检查. RHS的索引.

You can stop checking when the index of the LHS is greater or equal to the index of the RHS.

要将其转换为代码，

int is_palindrome(char const* s)
{
   size_t len = strlen(s);
   if ( len == 0 ) // An empty string a palindrome
   {
      return 1;
   }

   size_t i = 0;
   size_t j = len-1;
   for ( ; i < j; ++i, --j )
   {
      if ( s[i] != s[j] )
      {
         // the string is not a palindrome.
         return 0;
      }
   }

   // If we don't return from inside the for loop,
   // the string is a palindrome.
   return 1;
}

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