• 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 ;
    }
      
    

      

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  • 原文地址:https://www.cnblogs.com/NYNU-ACM/p/4237466.html
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