205. Isomorphic Strings

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.

Solution 1 Use a HashMap to keep the letter pairs 

29ms 43.31%

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

Solution 2 Actually using two arrays to keep the pair, the big array has is 512 (two times the size of ASCII), the former part to keep the s character, the latter part keeps the t character, give them a same value

17ms 83.90%

public class Solution {    public boolean isIsomorphic(String s, String t) {         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;    }}

This one uses two arrays and easier to understand

15ms 86.07%

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