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)

 1 public class Solution {
 2     public boolean isIsomorphic(String s, String t) {
 3         if (s == null || t == null)
 4             return false;
 5         if (s.length() != t.length())
 6             return false;
 7         if(s.length()==0 && t.length()==0)
 8             return true;
 9         int length = s.length();
10         Map<Character, Character> map = new HashMap<Character, Character>();
11         for (int i = 0; i < length; i++) {
12             char tmp1 = s.charAt(i);
13             char tmp2 = t.charAt(i);
14             if (map.containsKey(tmp1)) {
15                 if (map.get(tmp1) != tmp2)
16                     return false;
17             } else if (map.containsValue(tmp2)) {
18                 return false;
19             } else {
20                 map.put(tmp1, tmp2);
21             }
22         }
23         return true;
24     }
25 }

Solution 2

Use extra space to reduce time complexity.

 1 public class Solution {
 2     public boolean isIsomorphic(String s, String t) {
 3         if (s == null || t == null)
 4             return false;
 5         if (s.length() != t.length())
 6             return false;
 7         if(s.length()==0 && t.length()==0)
 8             return true;
 9         int length = s.length();
10         Map<Character, Character> map = new HashMap<Character, Character>();
11         Set<Character> counts = new HashSet<Character>();
12         for (int i = 0; i < length; i++) {
13             char tmp1 = s.charAt(i);
14             char tmp2 = t.charAt(i);
15             if (map.containsKey(tmp1)) {
16                 if (map.get(tmp1) != tmp2)
17                     return false;
18             } else if (counts.contains(tmp2)) {
19                 return false;
20             } else {
21                 map.put(tmp1, tmp2);
22                 counts.add(tmp2);
23             }
24         }
25         return true;
26     }
27 }