LeetCode Isomorphic Strings

原题链接在此: https://leetcode.com/problems/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.

Note:
You may assume both s and t have the same length.

题解：

这道题用两个Map建立s和t中对应Character的关系，一个是s到t的对应，一个是t到s的对应，要注意的是要双向检测两个Map里的对应值，否则就会出现下列错误：

Input:"ab", "aa"

Output:true

Expected:false

因为没有检查hm2，即使“aa”的第二个值 与 hm2中已有的a相同，也不会返回false。

Time Complexity: O(s.length()). Space: O(s.length()).

AC Java:

public class Solution {
    public boolean isIsomorphic(String s, String t) {
        if(s == null && t == null){
            return true;
        }
        if(s == null || t == null){
            return false;
        }
        if(s.length() != t.length()){
            return false;
        }
        
        Map<Character, Character> hm1 = new HashMap<Character, Character>();
        Map<Character, Character> hm2 = new HashMap<Character, Character>();
        int len = s.length();
        for(int i = 0; i<len; i++){
            if(hm1.containsKey(s.charAt(i)) && hm1.get(s.charAt(i)) != t.charAt(i)){
                return false;
            }
            if(hm2.containsKey(t.charAt(i)) && hm2.get(t.charAt(i)) != s.charAt(i)){
                return false;
            }
            hm1.put(s.charAt(i), t.charAt(i));
            hm2.put(t.charAt(i), s.charAt(i));
        }
        return true;
    }
}

很多key为char的map都可以用个长度为256的array表示.

思路和上面的相同, 两个map用扫到string的index位置联系在一起.

Time Complexity: O(s.length()). Space: O(1), 256 array 长度.

AC Java:

public class Solution {
    public boolean isIsomorphic(String s, String t) {
        if(s == null && t == null){
            return true;
        }
        if(s == null || t == null){
            return false;
        }
        if(s.length() != t.length()){
            return false;
        }
        
        int len = s.length();
        int [] m1 = new int[256];
        int [] m2 = new int[256];
        for(int i = 0; i<len; i++){
            if(m1[s.charAt(i)] != m2[t.charAt(i)]){
                return false;
            }
            m1[s.charAt(i)] = i+1;
            m2[t.charAt(i)] = i+1;
        }
        return true;
    }
}

 类似Word Pattern.