LeetCode Isomorphic Strings

原题链接在此: https://leetcode.com/problems/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.

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Hash Table

这道题用两个Map建立s和t中对应Character的关系，一个是s到t的对应，一个是t到s的对应，要注意的是要双向检测两个Map里的对应值，否则就会出现下列错误：

Input:"ab",
"aa"

Output:true

Expected:false

因为没有检查hm2，即使“aa”的第二个值 与 hm2中已有的a相同，也不会返回false。

Time Complexity: O(s.length()). Space: O(s.length()).

AC Java:

public class Solution {
public boolean isIsomorphic(String s, String t) {
if(s == null && t == null){
return true;
}
if(s == null || t == null){
return false;
}
if(s.length() != t.length()){
return false;
}

Map<Character, Character> hm1 = new HashMap<Character, Character>();
Map<Character, Character> hm2 = new HashMap<Character, Character>();
int len = s.length();
for(int i = 0; i<len; i++){
if(hm1.containsKey(s.charAt(i)) && hm1.get(s.charAt(i)) != t.charAt(i)){
return false;
}
if(hm2.containsKey(t.charAt(i)) && hm2.get(t.charAt(i)) != s.charAt(i)){
return false;
}
hm1.put(s.charAt(i), t.charAt(i));
hm2.put(t.charAt(i), s.charAt(i));
}
return true;
}
}