Isomorphic Strings 解答

Question

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 hashmap, remember to check both key and value. Time complexity O(n^2), space cost O(n).

hashMap.containsValue costs O(n)

public class Solution {
    public boolean isIsomorphic(String s, String t) {
        if (s == null || t == null)
            return false;
        if (s.length() != t.length())
            return false;
        if(s.length()==0 && t.length()==0)
            return true;
        int length = s.length();
        Map<Character, Character> map = new HashMap<Character, Character>();
        for (int i = 0; i < length; i++) {
            char tmp1 = s.charAt(i);
            char tmp2 = t.charAt(i);
            if (map.containsKey(tmp1)) {
                if (map.get(tmp1) != tmp2)
                    return false;
            } else if (map.containsValue(tmp2)) {
                return false;
            } else {
                map.put(tmp1, tmp2);
            }
        }
        return true;
    }
}

 Solution 2

Use extra space to reduce time complexity.

public class Solution {
    public boolean isIsomorphic(String s, String t) {
        if (s == null || t == null)
            return false;
        if (s.length() != t.length())
            return false;
        if(s.length()==0 && t.length()==0)
            return true;
        int length = s.length();
        Map<Character, Character> map = new HashMap<Character, Character>();
        Set<Character> counts = new HashSet<Character>();
        for (int i = 0; i < length; i++) {
            char tmp1 = s.charAt(i);
            char tmp2 = t.charAt(i);
            if (map.containsKey(tmp1)) {
                if (map.get(tmp1) != tmp2)
                    return false;
            } else if (counts.contains(tmp2)) {
                return false;
            } else {
                map.put(tmp1, tmp2);
                counts.add(tmp2);
            }
        }
        return true;
    }
}