Baozi Training Leetcode Solution 205: Isomorphic Strings

Blogger: https://blog.baozitraining.org/2019/07/leetcode-solution-205-isomorphic-strings.html

博客园:https://www.cnblogs.com/baozitraining/p/11112125.html

Problem Statement

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.

Example 1:

Input: s = "egg", t = "add"
Output: true

Example 2:

Input: s = "foo", t = "bar"
Output: false

Example 3:

Input: s = "paper", t = "title"
Output: true

Note:

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

Problem link

Video Tutorial

You can find the detailed video tutorial here

  • Youtube
  • B站

Thought Process

A relatively straightforward problem, very similar to Word Pattern. All we have to do is check the one to one mapping from string a to string b, also it needs to maintain a bijection mapping (meaning no two different characters in a should map to the same character in b)

Use bijection mapping

Check character one by one from a and b. If char in a hasn't been seen before, create a one to one mapping between this char in a and the char in b so later if this char in a is seen again, it has to map to b, else we return false. Moreover, need to make sure the char in b is never mapped by a different character.

Solutions

 1 public boolean isIsomorphic(String a, String b) {
 2         if (a == null || b == null || a.length() != b.length()) {
 3             return false;
 4         }
 5         Map<Character, Character> lookup = new HashMap<>();
 6         Set<Character> dupSet = new HashSet<>();
 7 
 8         for (int i = 0; i < a.length(); i++) {
 9             char c1 = a.charAt(i);
10             char c2 = b.charAt(i);
11 
12             if (lookup.containsKey(c1)) {
13                 if (c2 != lookup.get(c1)) {
14                     return false;
15                 }
16             } else {
17                 lookup.put(c1, c2);
18                 // this to prevent different c1s map to the same c2, it has to be a bijection mapping
19                 if (dupSet.contains(c2)) {
20                     return false;
21                 }
22                 dupSet.add(c2);
23             }
24         }
25         return true;
26     }

Time Complexity: O(N), N is the length of string a or string b

Space Complexity: O(N), N is the length of string a or string b because the hashmap and set we use

原文发布于微信公众号 - 包子铺里聊IT(baozitraining)

原文发表时间:2019-07-08

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