leetCode 205. Isomorphic Strings 哈希 字符串相似
2016-08-13 17:15
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205. Isomorphic Strings 字符串相似
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
You may assume both s and t have the same length.题目大意:
判断两个字符串是否相似。
思路:
使用双map来进行比较。map键为字符串元素,值为字符上一次出现的位置。
代码如下:
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.题目大意:
判断两个字符串是否相似。
思路:
使用双map来进行比较。map键为字符串元素,值为字符上一次出现的位置。
代码如下:
class Solution { public: bool isIsomorphic(string s, string t) { if(s.size() != t.size()) return false; unordered_map<char,int> maps; unordered_map<char,int> mapt; for(int i = 0;i < s.size();i++) { if(maps.find(s[i]) == maps.end() && mapt.find(t[i]) == mapt.end()) { maps.insert(pair<char,int>(s[i],i)); mapt.insert(pair<char,int>(t[i],i)); } else if(maps.find(s[i]) != maps.end() && mapt.find(t[i]) != mapt.end()) { if(maps[s[i]] != mapt[t[i]]) return false; else { maps[s[i]] = i; mapt[t[i]] = i; } } else return false; } return true; } };2016-08-13 17:15:21
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