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("\n\n");
}
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.\n\n",s) ;
else if(!Is_H_W(s)&&Is_J_X(s))
printf("%s -- is a mirrored string.\n\n",s) ;
else if(Is_H_W(s)&&!Is_J_X(s))
printf("%s -- is a regular palindrome.\n\n",s) ;
else
printf("%s -- is not a palindrome.\n\n",s) ;
}
return 0 ;
}
