LeetCode 题解(197) : Isomorphic Strings

题目：

Given two strings s and t, determine if they are isomorphic.

Two strings are isomorphic if the characters in s can be replaced to gett.

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.

题解：

HashTable + used数组。

C++版：

class Solution {public:    bool isIsomorphic(string s, string t) {        unordered_map<char, char> n;        vector<bool> used(95, false);        for(int i = 0; i < s.length(); i++) {            if(n.find(s[i]) != n.end()) {                if(n[s[i]] != t[i])                    return false;                continue;            }            if(used[t[i] - ' '])                return false;            n.insert(pair<char, char>(s[i], t[i]));            used[t[i] - ' '] = true;        }        return true;    }};

Java版：

import java.util.Hashtable;public class Solution {    public boolean isIsomorphic(String s, String t) {        Hashtable<Character, Character> m = new Hashtable<>();        boolean[] used = new boolean[95];        for(int i = 0; i < s.length(); i++) {            if(m.containsKey(s.charAt(i))) {                if(m.get(s.charAt(i)) != t.charAt(i))                    return false;                continue;            }            if(used[t.charAt(i) - ' '])                return false;            m.put(s.charAt(i), t.charAt(i));            used[t.charAt(i) - ' '] = true;        }        return true;    }}

Python版：

class Solution(object):    def isIsomorphic(self, s, t):        """        :type s: str        :type t: str        :rtype: bool        """        d = {}        used = [False] * 95        for i in range(len(s)):            if s[i] in d:                if d[s[i]] != t[i]:                    return False                continue            if used[ord(t[i]) - ord(" ")]:                return False            d[s[i]] = t[i]            used[ord(t[i]) - ord(" ")] = True        return True