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;
     }
 }