字符串的回文与镜像

给你一个字符串，如何判断这个字符串是不是回文串和镜像串。

看试一道很简单的题，但真正能把握住这道题却很难！

下面介绍三种方法，各有亮点：

　　第一种方法：先把可以镜像的字符用Hash表给存储起来，给出的字符串的一半入栈，如果这个字符串的长度为奇数，则这个字符串中间这个字符如果镜像后的字符和原来不相同，则这个字符串肯定不是镜像字符串，然后每次出栈一个，和下一个字符进行对比，如果出栈的字符和下一个字符不相等，则肯定不是回文字符串，如果在Hash表中，并镜像后，和下一个相等，则有可能是镜像字符串 ；

　　详细代码如下：

#include <cstdio>
#include <cstring>
#include <stack>
#include <map>
using namespace std;

int main() {
	char s[21];
	map<char, char> rev;
	rev['A'] = 'A';
	rev['H'] = 'H';
	rev['I'] = 'I';
	rev['E'] = '3';
	rev['J'] = 'L';
	rev['L'] = 'J';
	rev['M'] = 'M';
	rev['O'] = 'O';
	rev['T'] = 'T';
	rev['U'] = 'U';
	rev['V'] = 'V';
	rev['W'] = 'W';
	rev['X'] = 'X';
	rev['S'] = '2';
	rev['Y'] = 'Y';
	rev['Z'] = '5';
	rev['1'] = '1';
	rev['2'] = 'S';
	rev['3'] = 'E';
	rev['5'] = 'Z';
	rev['8'] = '8';

	while (scanf("%s", s) != EOF) {
		stack<char> checker;
		int len = strlen(s);

		bool palin = true, mirror = true;

		int pos = 0;
		for (; pos < len / 2; pos++) {
			checker.push(s[pos]);
		}
		if (len % 2 != 0) {
			if (!rev.count(s[pos]))
				mirror = false;
			if (rev[s[pos]] != s[pos])
				mirror = false;
			pos++;
		}

		for (; pos < len; pos++) {
			if (checker.top() != s[pos]) {
				palin = false;
			}
			if (!rev.count(checker.top()) || rev[checker.top()] != s[pos]) {
				mirror = false;
			}
			checker.pop();
		}

		printf("%s -- ", s);
		if (palin && mirror)
			printf("is a mirrored palindrome.");
		else if (palin && !mirror)
			printf("is a regular palindrome.");
		else if (!palin && mirror)
			printf("is a mirrored string.");
		else if (!palin && !mirror)
			printf("is not a palindrome.");
		printf("

");
	}

	return 0;
}

第二种方法：

　　先用Hash表将可以镜像的字符存储起来，从给出的字符串的最后一位开始逆序重建字符串，如果重建的和原来相同，就是回文，如果经过镜像之后重建的和原来相同，就是镜像串！

　　详细代码如下：

#include<iostream>
#include<algorithm>
#include<sstream>
#include<fstream>
#include<utility>
#include<cstdlib>
#include<cstring>
#include<string>
#include<bitset>
#include<vector>
#include<cstdio>
#include<cctype>
#include<cmath>
#include<queue>
#include<deque>
#include<stack>
#include<map>
#define ll long long
#define sc scanf
#define pf printf
#define pi 2*acos(0.0)
using namespace std;
int main()
{
    string s,a,b;
    char m[3000];

    memset(m,NULL,sizeof(m));
    m['A']='A';
    m['E']='3';
    m['H']='H';
    m['I']='I';
    m['J']='L';
    m['L']='J';
    m['M']='M';
    m['O']='O';
    m['S']='2';
    m['T']='T';
    m['U']='U';
    m['V']='V';
    m['W']='W';
    m['X']='X';
    m['Y']='Y';
    m['Z']='5';
    m['1']='1';
    m['2']='S';
    m['3']='E';
    m['5']='Z';
    m['8']='8';
    while(cin>>s){
    a=b="";
    int len=s.size();
    for(int i=len-1;i>=0;i--)
    {
        a+=s[i];
        b+=m[s[i]];
    }
    if (s==a && s==b)
        cout<<s<<" -- is a mirrored palindrome."<<endl<<endl;
    else if (s==a && s!=b)
        cout<<s<<" -- is a regular palindrome."<<endl<<endl;
    else if (s!=a && s==b)
        cout<<s<<" -- is a mirrored string."<<endl<<endl;
    else
        cout<<s<<" -- is not a palindrome."<<endl<<endl;
    }
    return 0;
}

 第三种方法：

#include<iostream>
#include<string.h>
#include<string>
#include<stdio.h>
using namespace std ;

int Is_H_W( char *s )	{
	int len = strlen(s) - 1 ;
	for(int i = 0 ; i < len ; i++,len--)
		if(s[i] != s[len])
			return 0 ;
	return 1 ;
}

char a[] = {'A',' ',' ',' ','3',' ',' ','H','I','L',' ','J','M',' ','O',' ',' ',' ','2','T','U','V','W','X','Y','5'} ;
char b[] = {'1','S','E',' ','Z',' ',' ','8',' '} ;

int Is_J_X( char *s )	{
	bool flag = false ;
	int len = strlen(s) - 1 ;
	if((len+1)%2)
		flag = true ;
	for(int i = 0 ; i <= len ; i++ , len--)	{
		if(s[i] >= 'A' && s[i] <= 'Z')	{
			if(a[s[i] - 'A'] != s[len])
				return 0 ;
		}
		else	{
			if(b[s[i] - '1'] != s[len])
				return 0 ;
		}
	}
	return 1 ;
}

int main()	{
	char s[22] ;
	while(gets(s))	{
		if(Is_H_W(s)&&Is_J_X(s))
			printf("%s -- is a mirrored palindrome.

",s) ;
		else if(!Is_H_W(s)&&Is_J_X(s))
			printf("%s -- is a mirrored string.

",s) ;
		else if(Is_H_W(s)&&!Is_J_X(s))
			printf("%s -- is a regular palindrome.

",s) ; 
		else
			printf("%s -- is not a palindrome.

",s) ; 
	}
	return 0 ;
}