Title Description

     Give a rule pattern And a string str , Judge str Follow the same rule .
     there follow Finger perfect match , for example , pattern Each letter and string in str There is a two-way connection between each non empty word in .

Example 1

 Input : pattern = "abba", str = "dog cat cat dog"
 Output : true

Example 2

 Input :pattern = "abba", str = "dog cat cat fish"
 Output : false

Example 3

 Input : pattern = "aaaa", str = "dog cat cat dog"
 Output : false

Tips :

• 1 <= pattern.length <= 300
• pattern Only lowercase letters
• 1 <= s.length <= 3000
• s Only lowercase letters and ’ ’
• s It doesn't contain Any leading or trailing pair of spaces
• s Every word in the is Single space Separate

Their thinking

C++ Using hash table to realize double mapping , That is, string pattern The elements in and str The words in , They are exchanged and stored in the hash table as keywords and values , Then you get two hash tables （A and B）, Then the comparison process , Make judgment in the process of comparison , If the current key appears , But if the values are different, it returns 0. The reader doesn't understand for a moment , The comparison process of the following two examples can be simulated manually on paper .

	 Ideas 
	for(int i=0;i<pattern.length();i++)
	{
    
		if( A.find( s pass the civil examinations i 's words  )!=A.end() && pattern pass the civil examinations i The letter of !=A[ s pass the civil examinations i 's words  ])//find Function to determine whether a key exists 
			return 0;
		if( B.find(pattern pass the civil examinations i The letter of )!=B.end() && s pass the civil examinations i 's words !=B[ pattern pass the civil examinations i The letter of  ])
			return 0;
	}

Example 1
Hashtable A

Hashtable B

Example 3
Hashtable A

Hashtable B

The code is as follows ：

// Rules of words 
bool wordPattern(string pattern, string s) {
    
	unordered_map<string,char>HashStr;
	unordered_map<char,string>Hashpattern;
	vector<string> ans;// Store string  s  The words in 
	string str="";
	for(int i=0;i<s.length();i++){
    
		if(s[i]!=' ')
			str+=s[i];
		else{
    
			ans.push_back(str);
			str="";	
		}
	}
	ans.push_back(str);	// here ans The elements in   Namely s The words in 
	
	if(pattern.length()!=ans.size())// Determine whether the number matches 
		return 0;
		
	for(int i=0;i<pattern.length();i++)// Exchange the elements of two strings as keys and values, respectively   Insert into hash table  
	{
    
		HashStr[ans[i]]=pattern[i];
		Hashpattern[pattern[i]]=ans[i];
	}
	
	for(int i=0;i<pattern.length();i++)// The essence is 
	{
    
		if(HashStr.find(ans[i])!=HashStr.end()&&pattern[i]!=HashStr[ans[i]])//find Function to query whether there exists 
			return 0;
		if(Hashpattern.find(pattern[i])!=Hashpattern.end()&&ans[i]!=Hashpattern[pattern[i]])
			return 0;
	}
	return 1;
}

This question mainly tests the operation of hash table , Readers can practice .