leetcode [290] - Word Pattern

 

Given a pattern and a string str, find if str 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 str.

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

Example 4:

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

Subject to the effect :

　　Given two strings str pattern and, in accordance with a given rule pattern, determines whether str also follow this rule.

Understanding:

　　See Example topics will judge misinterpreted as just this pattern is still congruent, this is wrong.

　　In fact, prior to this topic and a string of characters to replace similar topics. But, here is the word replacement character.

　　Save with character and word mapping relationship map. With the word save set has been mapped against different characters mapped to the same words.

　　Str first stored in the word dividing the vector, when the vector length and pattern of unequal size, false is returned.

　　Traversing pattern, the current character pattern [i] If it is not, it is determined whether there is a corresponding word has occurred is false; otherwise, the map is added to the map.

　　If the current character pattern [i] has occurred, to determine whether the word corresponding to the current word and the map maps are identical, continue to traverse, otherwise, it returns false.

Code C ++:

class Solution {
public:
    bool wordPattern(string pattern, string str) {
    Vector < String > splitWords;         // divided into words str 
    int the startPos = 0 , blankPos;         // record start position of the word, the position of the space 
    String Word;                     // store word 

    Map < char , String > m;             // store two mapping between string 
    Map < char , string > :: Iterator IT;
     SET < string > m_set;                 // prevent multiple characters mapped to the same word 
    int I = 0 ;                         // Traversal
    
    while ((blankPos = str.find(' ', startPos))!=str.npos) {
        Word = str.substr (the startPos, the startPos-blankPos);     // start position and length 
        splitWords.push_back (word);
        round as starting = blankPos + 1 ;
    }
    // process the last word 
    IF (the startPos < str.length ()) {
        word = str.substr(startPos);
        splitWords.push_back(word);
    }
    if (pattern.length() != splitWords.size())
        return false;
    while (pattern[i] != '\0') {
        it = m.find(pattern[i]);
        if (it == m.end()) {
            if (m_set.find(splitWords[i]) == m_set.end()) {
                m_set.insert(splitWords[i]);
            }
            else {
                return false;
            }
            m.insert(pair<char, string>(pattern[i], splitWords[i]));
            i++;
        }
        else {
            if (strcmp((it->second).c_str(), splitWords[i].c_str()) == 0)
                in ++ ;
            else 
                return  false ;
        }
    }
    return true;
}
};

operation result:

　　When execution: 4 ms, beat the 94.12% of all users to submit in C ++

　　Memory consumption: 8.9 MB, beat the 5.10% of all users to submit in C ++