题目描述

Given a pattern and a string s, find if s follows the same pattern.

Here follow means a full match, such that there is a bijection between a letter in pattern and a non-empty word in s.

Example 1：

Input: pattern = “abba”, s = “dog cat cat dog”
Output: true

Example 2:

Input: pattern = “abba”, s = “dog cat cat fish”
Output: false

Example 3:

Input: pattern = “aaaa”, s = “dog cat cat dog”
Output: false

Example 4:

Input: pattern = “jquery”, s = “jquery”
Output: false

Constrains:

• 1 <= pattern.length <= 300
• pattern contains only lower-case English letter.
• 1 <= s.length <= 3000
• s contains only lowercase English letters ans spaces ’ '.
• s does not contain any leading or trailing spaces.

解答思路

这题需要s包含的单词与pattern逐字母的进行映射，比如pattern是abba，那么s包含的字符也需要是对称的。这里需要注意当pattern与s一样的时候，输出为false，因为s应该与pattern每一个字母做映射，而不是整体。
思路类似205. Isomorphic String，做双向的映射。pattern每个字母与s里的每个单词做映射，单向的还不够，因为有可能出现abc，与"dog cat dog"对应时，a和c都对应上dog了。

class Solution {
public:bool wordPattern(string pattern, const string& s) {       return wordPatternHelp(pattern, s) && wordPatternHelp2(pattern, s);}void split_string(vector<string> &ves, string str){istringstream ss(str);string s;while(ss >> s){ves.push_back(s);}}bool wordPatternHelp(string pattern, string s) {unordered_map<char, string> m;vector<string> ves;split_string(ves, s);if(pattern.length() > ves.size())return false;for(int i = 0; i < pattern.length(); i++){if(m.find(pattern[i]) != m.end()){if(m[pattern[i]] != ves[i])return false;}elsem[pattern[i]] = ves[i];}return true;}bool wordPatternHelp2(string pattern, string s){unordered_map<string, char> m;vector<string> ves;split_string(ves, s);for(int i = 0; i < ves.size(); i++){if(m.find(ves[i]) != m.end()){if(m[ves[i]] != pattern[i])return false;}elsem[ves[i]] = pattern[i];}return true;}
};