Given two strings s andt, 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 "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.
题意：判断两个字符串同形字符串，就是说，将s中的字符替换为t中的相应字符，能够保证s中不同的字符替换成t中的不同的字符，可以想象成一个加密过程，利用某种映射，将s加密成t
public class Solution {public boolean isIsomorphic(String s, String t) {//利用map保存映射关系Map<Character,Character> map = new HashMap<Character,Character>();for(int i = 0 ; i < s.length() ; i ++){char key = s.charAt(i);char value = t.charAt(i);if(map.containsKey(key)){if(value != map.get(key)){//存在一条key->x的映射，且x != valuereturn false;}}else if(map.containsValue(value)){return false;//已经存在一条x->value的映射,但是x != key}else{//不存在key->x映射和x->value的映射map.put(key,value);}}return true;}}搜了一下，感觉别人的算法跟我的差不多，没找到很高效的算法，，为什么我的运行时间那么慢
更重要的是心理上的完全自由和放松，
