Leetcode 205. Isomorphic Strings
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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.
s思路:
1. isomorphic表示同质,对string来说,就是结构相同,里面的字母是谁不重要,比如:abb和cdd就是isomorphic,因为形式上是一样的。如何判断呢?
2. 略微思考,有点没有眉目,开始有点细微的紧张。不过还好,做得多了,有底气了,和这些情绪都相处得很好。回到正题,两个string同时遍历,例如:”paper”, “title”,把p映射成t,a映射成i,第二次遇到p时,就查询之前p映射的是多少,和现在打算映射的值是否一致。如果能通过所有一致性检查,说明就是isomorphic的;中间任何地方不一致,就不是!
3. 代码第一次写的时候,默认只用一个unordered_map来存s[i]->t[i]的映射,但是对”agg”和”ggg”则不正确了,还需要存t[i]->s[i]的映射,因为只有这个映射,a和g映射成g都是合法的,但题目要求,不允许两个字母映射成一个字母。因此需要两个map来保存相互的映射关系最保险!
class Solution {public: bool isIsomorphic(string s, string t) { // unordered_map<char,char> mm1,mm2; int i=s.size(); for(int i=0;i<s.size();i++){ if(mm1.count(s[i])){ if(mm1[s[i]]!=t[i]) return false; }else if(mm2.count(t[i])){ if(mm2[t[i]]!=s[i]) return false; } else{ mm1[s[i]]=t[i]; mm2[t[i]]=s[i]; } } return true; }};
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