Isomorphic Strings

package cn.edu.xidian.sselab.hashtable;

import java.util.HashMap;
import java.util.Map;

/**
 *
 * @author zhiyong wang
 * title:
 * content:
 * 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.
 *
 */
public class IsomorphicStrings {

    //自己想到的方法，注意else里面多加了一个条件，防止出现一对多的映射
    public boolean isIsomorphic(String s, String t){
        if(s == null || t == null) return false;
        int lengthS = s.length();
        int lengthT = t.length();
        if(lengthS != lengthT) return false;
        if(lengthS == 0 && lengthT != 0 || lengthS !=0 && lengthT == 0) return false;
        Map index = new HashMap();
        for(int i=0;i<lengthS;i++){
            if(index.containsKey(s.charAt(i))){
                if(!index.get(s.charAt(i)).equals(t.charAt(i)))
                    return false;
            }else{
                if(index.containsValue(t.charAt(i)))
                    return false;
                index.put(s.charAt(i), t.charAt(i));
            }
        }
        return true;
    }
    
    //参考了大牛的做法,这个题与Valid Vnagram解题思路类似，但又不同,这里不是从0开始加，而是循环次数加1
     public boolean isIsomorphics(String s, String t) {
         int[] temp1 = new int[256];
         int[] temp2 = new int[256];
         int length = s.length();
         for(int i=0;i<length;i++){
             if(temp1[s.charAt(i)] != temp2[t.charAt(i)])
                 return false;
             temp1[s.charAt(i)] = i+1;
             temp2[t.charAt(i)] = i+1;
         }
          return true;
        }
    
}