#leetcode#Isomorphic Strings-优快云博客

本文介绍了一种算法，用于判断两个字符串是否为同构字符串。即一个字符串中的字符是否可以替换得到另一个字符串，要求所有出现的字符都必须用另一个字符替换，并保持字符顺序不变。

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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 Solution {
    public boolean isIsomorphic(String s, String t) {
        if(s == null && t == null)
            return true;
        if(s ==  null || t == null)
            return false;
        
        Map<Character, Character> map1 = new HashMap<>();
        Map<Character, Character> map2 = new HashMap<>();
        
        for(int i = 0; i < s.length(); i++){
            char sChar = s.charAt(i);
            char tChar = t.charAt(i);
            
            if(!map1.containsKey(sChar)){
                map1.put(sChar, tChar);
            }else{
                if(tChar != map1.get(sChar)){
                    return false;
                }
            }
            
            if(!map2.containsKey(tChar)){
                map2.put(tChar, sChar);
            }else{
                if(sChar != map2.get(tChar)){
                    return false;
                }
            }
        }
        return true;
    }
}

注意：要求双向对应，比如 ‘ab’ 对应 ‘aa’应该返回false，所以用了两个hashmap