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.

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

Solution1: time complexity is O(n^2), space complexity is O(n).

public class Solution {
    public boolean isIsomorphic(String s, String t) {
        if (s == null && t == null || (s.length() == 0 && t.length() == 0)) {
            return true;
        }
        
        if (s.length() != t.length()) {
            return false;
        }
        
        HashMap map = new HashMap();
        for (int i = 0; i < s.length(); i++) {
            char cs = s.charAt(i);
            char ct = t.charAt(i);
            
            Character key = getKey(map, ct);
            if (key != null && key != cs) {   // deal with case: ab -> aa, different keys have same values
                return false;
            }
            
            if (map.containsKey(cs)) {
                if (ct != map.get(cs)) {
                    return false;
                }
            } else {
                map.put(cs, ct);
            }
        }
        return true;
    }
    
    public Character getKey(HashMap map, Character target) {
        for (Map.Entry entry : map.entrySet()) {
            if (entry.getValue().equals(target)) {
                return entry.getKey();
            }
        }
        return null;
    }
}

Reference: http://www.programcreek.com/2014/05/leetcode-isomorphic-strings-java/

Solution2: time complexity is O(n ^ 2), space complexity is O(n). 
Time complexity of HashMap.containsValue() is O(n).

public class Solution {
    public boolean isIsomorphic(String s, String t) {
        if (s == null && t == null || (s.length() == 0 && t.length() == 0)) {
            return true;
        }
        
        HashMap map = new HashMap();
        for (int i = 0; i < s.length(); i++) {
            char cs = s.charAt(i);
            char ct = t.charAt(i);
            
            if (map.containsKey(cs)) {
                if (map.get(cs) != ct) {
                    return false;
                } else {
                    continue;
                }
            } else if (map.containsValue(ct)) {
                return false;
            }
            map.put(cs, ct);
        }
        return true;
    }
}