lintcode(638)Strings Homomorphism

描述：

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.

 注意事项

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

样例：

Given s = "egg", t = "add", return true.

Given s = "foo", t = "bar", return false.

Given s = "paper", t = "title", return true.

思路：

判断对应位置字符是否成唯一映射，同时判断是否出现重复映射。

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        HashMap<Character , Character> record = new HashMap<>();        HashSet<Character> repeat = new HashSet<>();        for(int i = 0;i<s.length();i++){            if(!record.containsKey(s.charAt(i))){                if(repeat.contains(t.charAt(i))){                    return false;                }                record.put(s.charAt(i) , t.charAt(i));                repeat.add(t.charAt(i));            }else{                if(record.get(s.charAt(i)) != t.charAt(i)){                    return false;                }            }        }        return true;    }}