[英]Check if a string is palindrome in C
i've a question about this code i'm writing for an exercise. 我对我正在练习的这段代码有疑问。 I've to check if a string is palindrome.
我要检查一个字符串是否是回文。 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!
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!
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;
}
Your function to determine whether a string is a palindrome is not well thought out. 判断字符串是否为回文符的功能未得到充分考虑。
Let's say you have a string s
of length l
. 假设您有一个长度为
l
的字符串s
。 The characters in the string are laid out as: 字符串中的字符布局为:
Indices: 0 1 2 3 l-4 l-3 l-2 l-1
+----+----+----+----+- ... -+----+----+----+----+
| | | | | ... | | | | |
+----+----+----+----+- ... -+----+----+----+----+
If the string is a palindrome, 如果字符串是回文,
s[0] = s[l-1]
s[1] = s[l-2]
...
You can stop checking when the index of the LHS is greater or equal to the index of the RHS. 当LHS的索引大于或等于RHS的索引时,您可以停止检查。
To translate that into code, 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;
}
MARCO try this. MARCO试试这个。
#include <stdio.h>
#include <stdlib.h>
#include<string.h>
char* cargar (char*);
int pali (char*);
int main()
{
char*texto=NULL;
texto=cargar(texto);
int res=pali(texto);
if(res==strlen(texto)){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* a)
{
int flag=0,i;
int len=strlen(a);
for (i=0;i<len;i++)
{
if(a[i]==a[len-i-1])
flag=flag+1;
}
return flag;
}
You pali
function tests if the first character of the string is equal to the last character and then invokes itself for a position of the second character of the string. 您的
pali
函数测试字符串的第一个字符是否等于最后一个字符,然后调用自身以查找字符串的第二个字符。 Note however, it does not modify the end of a string , so the recursive invocation compares the second character again to the last one. 但是请注意,它不会修改字符串的结尾 ,因此递归调用会将第二个字符再次与最后一个字符进行比较。 Then you compare the third charcter to the last one... Finaly
pali
returns 1
if all characters are equal to the last one, that is if all are equal. 然后,您将第三个字符与最后一个字符进行比较...如果所有字符都等于最后一个字符 ,也就是说,如果所有字符都相等,则Finaly
pali
返回1
。
Try this: 尝试这个:
int pali (char* texto)
{
char* end;
for(end = texto; *end != '\0'; end ++)
;
for(--end; texto < end; ++texto, --end) {
if(* texto != * end)
return 0;
}
return 1;
}
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