NOJ Palindromes

Palindromes

时间限制(普通/Java) : 1000 MS/ 3000 MS 运行内存限制 : 81920 KByte
总提交 : 65 测试通过 : 36

题目描述

Write a program that determines if each input string is a palindrome. A palindrome is a

string that reads exactly the same in both forward and reverse directions. For something

to be considered a palindrome, it must be at least 1 character long. For the purposes of

your program, ignore any characters that are not letters, as well as spaces when

determining if a string is a palindrome.

输入

The first line of input contains an integer N that indicates the number of test strings to

follow. On each subsequent line there will be a single test string. Here is a sample:

输出

For each test string, output "yes" if the string was a palindrome, and "no" if it was not a

palindrome. Remember: Ignore any characters that are not letters, as well as spaces.

样例输入

5
able ##was I, e****re I s.aw $Elba
this is not a palindrome
A man, a plan, a canal, Panama
another random string
Sator Arepo Tenet Opera Rotas

样例输出

yes
no
yes
no
yes

分析：判断输入的字符串是否回文，仅考虑字符串中的字母，我们可以将输入字符串中的字母存在另一个数组里进行判断。

实现代码：

<span style="font-size:12px;">#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int T;
int main()
{
  //  freopen("data.in","r",stdin);
    scanf("%d",&T);
    getchar();
  lp:while(T--)
    {
        char a[200000],b[200000];
        a[0]='';
        b[0]='';
        gets(a);
        int cnt=0;
        for(int i=0;i<strlen(a);i++)
        {
            if((a[i]>='A'&&a[i]<='Z')||(a[i]>='a'&&a[i]<='z'))
                b[cnt++]=a[i];
        }
        for(int i=0,j=cnt-1;i<=j;i++,j--)
        {
            if(b[i]!=b[j]&&b[i]!=b[j]+32&&b[i]+32!=b[j])
            {
                printf("no
");
                goto lp;
            }
        }
        printf("yes
");
    }
}</span>