回文写法

测试字符串是否是回文字串

   1. bool is_palindrome(char *str, int size)  
   2. {  
   3.    int tmp1 = size-2, tmp2 = (size-1)/2;  
   4.    for (int i = 0; i < tmp2; ++i)  
   5.    {  
   6.        if (str[i] != str[tmp1-i])  
   7.          return false;  
   8.    }  
   9.    return true;  
  10. }  

然后，不管对应位置的大小写：

可以这么写：

# bool is_palindrome(char *str, int size)  
# {  
#    int tmp1 = size-2, tmp2 = (size-1)/2;  
#    for (int i = 0; i < tmp2; ++i)  
#    {  
#        int char_dif = str[i]-str[tmp1-i];  
#        if (str[i] == str[tmp1-i] || abs(char_dif)==32)  
#          continue;  
#        return false;  
#    }  
#    return true;  
# } 

忽略源字符串内的标点符号，测试是否是回文串：

# #include <iostream>  
# #include <iostream>  
# #include <math.h>  
# using namespace std;  
#   
#   
# struct res  
# {  
#     char *s;  
#     int size;  
# };  
# res ignore_punctuate(char *str, int size)  
# {  
#   int tmp = size-2;  
#   char *str1 = new char[size];  
#   int j = 0;  
#   for (int i = 0; i <= tmp; ++i)  
#   {  
#      if (!ispunct(str[i]))  
#      {  
#          str1[j++]=str[i];  
#      }  
#   }  
#   str1[j]='\0';  
#   res tmp1;  
#   tmp1.s = str1;  
#   tmp1.size=j+1;  
#   return tmp1;  
# }  
# bool is_palindrome(char *str, int size)  
# {  
#     int tmp1 = size-2, tmp2 = (size-1)/2;  
#     for (int i = 0; i < tmp2; ++i)  
#     {  
#         int char_dif = str[i]-str[tmp1-i];  
#         if (str[i] == str[tmp1-i] || abs(char_dif)==32)  
#             continue;  
#         return false;  
#     }  
#     return true;  
# }  
#   
# int main()  
# {  
#     char a[]="abb,A";  
#     int size = sizeof(a)/sizeof(*a);  
#     res tmp = ignore_punctuate(a,size);  
#     cout << is_palindrome(tmp.s, tmp.size) << endl;  
# }