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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 "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.
给定两个字符串,判断它们是否属于同构词。我们借助一个map和一个set,map用来存放两个字符串中对应字符的关系,set用来存放map中的value。题目假设两个字符串长度相同,假设我们比较到了第i个字符,如果map中已经存放了key为s.chatAt(i)的键值对,我们就比较t.charAt(i)与get(s.charAt(i))是否相等,如果不等,就返回false,如果相等就继续搜索。如果map中没有存放key为s.charAt(i)的键值对,我们此时要判断map中的value是否有t.charAt(i),如果set中已经存在t.charAt(i), 直接返回false;否则将这一个键值对放入map中,并且把value放入set中。代码如下:
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.
给定两个字符串,判断它们是否属于同构词。我们借助一个map和一个set,map用来存放两个字符串中对应字符的关系,set用来存放map中的value。题目假设两个字符串长度相同,假设我们比较到了第i个字符,如果map中已经存放了key为s.chatAt(i)的键值对,我们就比较t.charAt(i)与get(s.charAt(i))是否相等,如果不等,就返回false,如果相等就继续搜索。如果map中没有存放key为s.charAt(i)的键值对,我们此时要判断map中的value是否有t.charAt(i),如果set中已经存在t.charAt(i), 直接返回false;否则将这一个键值对放入map中,并且把value放入set中。代码如下:
public class Solution { public boolean isIsomorphic(String s, String t) { Map<Character, Character> hm = new HashMap<Character, Character>(); Set<Character> set = new HashSet<Character>(); for(int i = 0; i < s.length(); i++) { if(hm.containsKey(s.charAt(i))) { if(hm.get(s.charAt(i)) != t.charAt(i)) return false; } else { if(set.contains(t.charAt(i))) { return false; } else { hm.put(s.charAt(i), t.charAt(i)); set.add(t.charAt(i)); } } } return true; } }
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