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isIsomorphic

Posted on 2015-06-03 00:18  cKK  阅读(215)  评论(0编辑  收藏  举报

超时版:

/*
Given two strings s and t, determine if they are isomorphic.

Two strings are isomorphic if the characters in s can be replaced to get t.

All occurrences of a character must be replaced with another character while preserving the order of characters. No two characters may map to the same character but a character may map to itself.

For example,
Given "egg", "add", return true.

Given "foo", "bar", return false.

Given "paper", "title", return true.

Note:
You may assume both s and t have the same length.*/
import java.util.*;

public class Isomorphic_strings {

    public static void main(String[] args)
    // TODO Auto-generated method stub
    {
        System.out.println(isIsomorphic("papera", "titlei"));
        /*
         * String str="dauqka"; String str1=str.replace('a','x'); str=str1;
         * System.out.println(str);
         */

    }

    public static boolean isIsomorphic(String s, String t) {

        int x = 0;
        if (s.length() != t.length())
            return false;
        char[] s_chs = s.toCharArray();
        char[] t_chs = t.toCharArray();
        for (int i = 0; i < s_chs.length; i++) {
            if (!(s_chs[i] == t_chs[i])) {
                for (int j = 0; j < i; j++) {
                    if ((t_chs[j] == t_chs[i]))
                        x = 1;
                }
            }
            if (x == 0)
                t = t.replace(t_chs[i], s_chs[i]);
            if (x == 1) {
                t_chs[i] = s_chs[i];
                t = String.valueOf(t_chs);

            }
            t_chs = t.toCharArray();
        }

        return (s.equals(t));

    }

}

 AC

public static boolean isIsomorphic(String s, String t) {

        
        Map<Character, Character> hm = new HashMap<Character, Character>();
        if (s.length() != t.length())
            return false;
        char[] s1 = s.toCharArray();
        char[] t1 = t.toCharArray();
        for (int i = 0; i < s1.length; i++) {
            if (hm.containsKey(s1[i])) {
                if ((t1[i] != hm.get(s1[i])))
                    return false;
            }
            hm.put(s1[i], t1[i]);
        }

        return true;
    }

}