LeetCode_Isomorphic Strings

Isomorphic Strings

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.

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.

判断两个字符串是否同构：s中的每个字符按顺序映射到t中一个字符，映射关系不能是一对多，例如"foo"和"bar"，按顺序映射：f->b，o->a，o->r，o同时对a和r，所以foo和egg不是同构的；同理，不能多对一，即s中不同字符不能对应t中相同的字符，例如，"ab"和"aa"，a->a，b->a，a和b同时对应a，所以ab和aa不符合条件。

以s中的每个字符作为key，t中的每个字符作为value，利用Java中的map容器做映射。

public class Solution {
    public boolean isIsomorphic(String s, String t) {
    Map<Character,Character> mp1 = new HashMap<>();
		final int len1 = s.length();
		final int len2 = s.length();
		if(len1!=len2) return false;
		if(len1==0) return true;
		for(int i = 0;i < len1;i++)
		{
			if(!mp1.containsKey(s.charAt(i)))
			{
				mp1.put(s.charAt(i),t.charAt(i));
				for(int j = 0;j<i;j++)
				{
					if(mp1.get(s.charAt(i))==mp1.get(s.charAt(j)))//多对一
					{
						return false;
					}
				}
			}
			else
			{
				if(mp1.get(s.charAt(i))!=t.charAt(i))//一对多
				{
					return false;
				}
			}
		}
        return true;
    }
}

时间复杂度较高，期待更高效的方法。