lintcode :同构字符串

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.

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解题

定义HashMap 

public class Solution {
    public boolean isIsomorphic(String s, String t) {
        if(s==null || t ==null)
            return false;
        if(s.length()!=s.length())
            return false;
        if(s==null && t == null)
            return true;
        HashMap<Character,Character> map = new HashMap<Character,Character>();
        for(int i = 0;i< s.length();i++){
            char c1 = s.charAt(i);
            char c2 = t.charAt(i);
            Character c = getKey(map,c2);
            if(c !=null && c!=c1)
                return false;
            else if(map.containsKey(c1)){
                if(c2!=map.get(c1))
                    return false;
            }else{
                map.put(c1,c2);
            }
        }
        return true;
    }
 
    public Character getKey(HashMap<Character,Character> map,Character target){
        for(Map.Entry<Character,Character> entry:map.entrySet()){
            if(entry.getValue().equals(target)){
                return entry.getKey();
            }
        }
        return null;
    }
}

或者

public class Solution {
    public boolean isIsomorphic(String s, String t) {
        if(s==null || t ==null)
            return false;
        if(s.length()!=s.length())
            return false;
        if(s==null && t == null)
            return true;
        HashMap<Character,Character> map = new HashMap<Character,Character>();
        for(int i = 0;i< s.length();i++){
            char c1 = s.charAt(i);
            char c2 = t.charAt(i);
            Character c = map.get(c1);
            // key c1 value c2
            if(map.containsKey(c1)){
                 if(map.get(c1).equals(c2))
                     continue;
                 else
                     return false;
            }else{
                if(!map.containsValue(c2))
                    map.put(c1,c2);
                else
                    return false;
 
            }
 
        }
        return true;
    }
 
}

这里的字符串都是字母，前256个字符，将其映射到后256个字符

public class Solution {
    public boolean isIsomorphic(String s, String t) {
        if(s==null || t ==null)
            return false;
        if(s.length()!=s.length())
            return false;
        if(s==null && t == null)
            return true;
        int[] m = new int[512];
        for (int i = 0; i < s.length(); i++) {
            if (m[s.charAt(i)] != m[t.charAt(i)+256])
                return false;
            m[s.charAt(i)] = m[t.charAt(i)+256] = i+1;
        }
        return true;
    }
 
}