LeetCode OJ 205 Isomorphic Strings
2015-05-03 13:49
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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
Given
Given
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
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
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.
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 != value return 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
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