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.

Notice

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

Example

Given s = "egg", t = "add", return true.
Given s = "foo", t = "bar", return false.
Given s = "paper", t = "title", return true.

思路

用HashMap记录s与t每个字符对应关系
用HashSet记录t中的每个字符是否已经被匹配过

Conor Case: s与t如果长度不等，则直接返回false

1. 遍历s和t，分别同时取s和t对应位置的字符，判断mapping是否包含s对应的字符， 如果不包含s对应的字符：
   1）如果对应位置的t字符也不在set中（说明并未出现过），那么则向mapping中加入s和t对应字符，再向set中加入t对应字符
   2）如果set中已经出现了对应位置的t字符。换句话，s没有出现过，但是t出现过了，比如s = 'aa', t = 'bc', 则说明t没有遵守pattern，返回false。

2. 如果包含s对应的字符（说明s对应字符已经出现）：
   1）如果mapping.get(s.charAt(i)) 不等于t.charAt(i)， 则说明t没有遵守pattern，返回false。

   2. 如果相等，说明遵守，什么也不做，继续。

相关题目

Word Pattern （http://www.jianshu.com/p/5fcef82f11ef）
Word Pattern II

public class Solution {
    /*
     * @param s: a string
     * @param t: a string
     * @return: true if the characters in s can be replaced to get t or false
     */
    public boolean isIsomorphic(String s, String t) {
        // write your code here
        if (s == null || t == null ||s.length() != t.length()) {
            return false;
        }
        HashMap<Character, Character> mapping = new HashMap<>();
        Set<Character> set = new HashSet<Character>();
        
        for (int i = 0; i < s.length(); i++) {
            if (!mapping.containsKey(s.charAt(i))) {
                if (set.contains(t.charAt(i))) {
                    return false;
                }
                mapping.put(s.charAt(i), t.charAt(i));
                set.add(t.charAt(i));
            } else {
                if (mapping.get(s.charAt(i)) != t.charAt(i)) {
                    return false;
                }
            }
        }
        return true;
    }
}