指针回文(有人可以通过简单的方法逐步解释以下代码...非常感谢！) [英] pointer palindrome(can someone please explain the following code step by step and in a easy way...thanks a lot!!!)

问题描述

#include <stdio.h>
#include <ctype.h>
#define STR_LEN 80

int main()
{
    char s[STR_LEN+1];
    int i = 0;
    char *left;
    char *right;
    char *end;

    for (i = 0; i< = STR_LEN; i++) 
    {
        s[i] = ' ';
    }
    printf("Enter: ");
    fgets(s, STR_LEN+1, stdin);

    left = &s[0];
    end = right = &s[STR_LEN];

    while (left<right) 
    {
        while ((!((*left>=65)&&(*left<=90)))&&
               (!((*left>=97)&&(*left<=122)))&&
               (left<end)) 
        {
            left++;
        }
        while ((!((*right>=65)&&(*right<=90)))&&
               (!((*right>=97)&&(*right<=122)))&&
               (right>s)) 
        {
            right--;
        }
        if (left == end) 
        {
            break;
        }
        if (tolower(*left) != tolower(*right)) 
        {
            printf("\nNo Palindrome!\n\n");
            return 0;
        }
        left++; 
        right--;
  }

  printf("\nPalindrome!\n\n");
  return 0;
}

推荐答案

它从字符数组的每个末端开始，以查找字符匹配项.给定字符串"abccba"，它将比较位置0和5、1、4和2和3处的字符是否相同.如果是这样，那就是回文.如果碰巧不匹配，那就不是回文.从技术上讲，它不应该区分大小写，但这只是实现的问题.

顺便说一句，我对格式进行了一些更改，以便更轻松地了解其工作原理.

It''s starting at each end of the character array, looking for character matches. Given the string "abccba", it''s comparing the character at position 0 and 5, 1 and 4, and 2 and 3 to see if they''re the same. If they are, it''s a palindrome. If it runs into a pair that doesn''t match, it''s not a palindrome. Technically, it should not be case-sensitive, but that''s a matter of implementation.

BTW, I change the formatting a little so it''s easier to see how it''s working.

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