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.
Approach #1: C++.
class Solution { public: bool isIsomorphic(string s, string t) { int len = s.size(); int ArrS[256] = {0}; int ArrT[256] = {0}; for (int i = 0; i < len; ++i) { if (charArrS[s[i]] != charArrT[t[i]]) { return false; } ArrS[s[i]] = i + 1; ArrT[t[i]] = i + 1; } return true; } };
Approach #2: Java.
class Solution { public: bool isIsomorphic(string s, string t) { int len = s.size(); int ArrS[256] = {0}; int ArrT[256] = {0}; for (int i = 0; i < len; ++i) { if (charArrS[s[i]] != charArrT[t[i]]) { return false; } ArrS[s[i]] = i + 1; ArrT[t[i]] = i + 1; } return true; } };
Approach #3: Python.
class Solution(object): def isIsomorphic(self, s, t): """ :type s: str :type t: str :rtype: bool """ arrS = {} arrT = {} for i, val in enumerate(s): arrS[val] = arrS.get(val, []) + [i] for i, val in enumerate(t): arrT[val] = arrT.get(val, []) + [i] return sorted(arrS.values()) == sorted(arrT.values())
Time Submitted | Status | Runtime | Language |
---|---|---|---|
a few seconds ago | Accepted | 236 ms | python |
5 minutes ago | Accepted | 8 ms | java |
13 minutes ago | Accepted | 4 ms | cpp |
Analysis: