leetcode 205. Isomorphic Strings

题目

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.

理解

判断两个字符串是否是同分异构，也就是判断两个字符串里面的字符是不是一一对应。

解决

两个map

正向保证不会一个key对应两个value，反向保证不会两个key对应同一个value

public class Solution {
    public boolean isIsomorphic(String s, String t) {
        if(s==null && t==null){
            return true;
        }else if(s==null || t==null){
            return false;
        }
        if(s.length() != t.length()){
            return false;
        }
        int N = s.length();
        Map<Character,Character> map = new HashMap<Character,Character>();
        Map<Character,Character> reservedMap = new HashMap<Character,Character>();
        for(int i=0; i<N; i++){
            if(map.containsKey(s.charAt(i))){
                if(map.get(s.charAt(i))!=t.charAt(i)){
                    return false;
                }
            }else{
                map.put(s.charAt(i),t.charAt(i));
            }
            if(reservedMap.containsKey(t.charAt(i))){
                if(reservedMap.get(t.charAt(i))!=s.charAt(i)){
                    return false;
                }
            }else{
                reservedMap.put(t.charAt(i),s.charAt(i));
            }
        }
        return true;
    }
}

一个map，一个set

map保证不会一个key对应两个value，set保证不会两个key对应同一个value

public class Solution {
    public boolean isIsomorphic(String s, String t) {
        if(s==null && t==null){
            return true;
        }else if(s==null || t==null){
            return false;
        }
        if(s.length() != t.length()){
            return false;
        }
        int N = s.length();
        Map<Character,Character> map = new HashMap<Character,Character>();
        Set<Character> set = new HashSet<Character>();
        for(int i=0; i<N; i++){
            if(map.containsKey(s.charAt(i))){
                if(map.get(s.charAt(i))!=t.charAt(i)){
                    return false;
                }
            }else if(set.contains(t.charAt(i))){
                return false;
            }else{
                map.put(s.charAt(i),t.charAt(i));
                set.add(t.charAt(i));
            }
        }
        return true;
    }
}