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.
题目很简单,也很容易想到方法,就是记录遍历s的每一个字母,并且记录s[i]到t[i]的映射,当发现与已有的映射不同时,说明无法同构,直接return false。但是这样只能保证从s到t的映射,不能保证从t到s的映射,所以交换s与t的位置再重来一遍上述的遍历就OK了。
1 class Solution { 2 public: 3 bool isIsomorphic(string s, string t) { 4 if (s.length() != t.length()) return false; 5 map<char, char> mp; 6 for (int i = 0; i < s.length(); ++i) { 7 if (mp.find(s[i]) == mp.end()) mp[s[i]] = t[i]; 8 else if (mp[s[i]] != t[i]) return false; 9 } 10 mp.clear(); 11 for (int i = 0; i < s.length(); ++i) { 12 if (mp.find(t[i]) == mp.end()) mp[t[i]] = s[i]; 13 else if (mp[t[i]] != s[i]) return false; 14 } 15 return true; 16 } 17 };