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.
秒小题好爽的说。
思路:保存两个字符串字母对应的字母,出现不一致就false
我的代码:
bool isIsomorphic(string s, string t) { unordered_map<char, char> map1, map2; for(int i = 0; i < s.size(); ++i) { if(map1.find(s[i]) == map1.end() && map2.find(t[i]) == map2.end()) { map1[s[i]] = t[i]; map2[t[i]] = s[i]; } else if(map1[s[i]] != t[i] || map2[t[i]] != s[i]) return false; else; } return true; }
网上C版本代码 免去了查找 更快
bool isIsomorphic(char* s, char* t) { char mapST[128] = { 0 }; char mapTS[128] = { 0 }; size_t len = strlen(s); for (int i = 0; i < len; ++i) { if (mapST[s[i]] == 0 && mapTS[t[i]] == 0) { mapST[s[i]] = t[i]; mapTS[t[i]] = s[i]; } else { if (mapST[s[i]] != t[i] || mapTS[t[i]] != s[i]) return false; } } return true; }