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
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.
You may assume both s and t have the same length.
Solution: Hash table. The test cases assume ASCII characters. A valid mapping is indeed a bijection. The easiest way to determine whether such a mapping exists is to test both directions. The beauty of symmetry!
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 | public class Solution { public boolean isIsomorphicOneWay(String s, String t) { char[] map = new char[256]; Arrays.fill(map, '*'); for (int i = 0; i < s.length(); i++) { int idx = (int)s.charAt(i); if (map[idx] == '*') { map[idx] = t.charAt(i); } else { if (map[idx] != t.charAt(i)) { return false; } } } return true; } public boolean isIsomorphic(String s, String t) { if (s.length() != t.length()) { return false; } return isIsomorphicOneWay(s, t) && isIsomorphicOneWay(t, s); } } |
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