【一天一道LeetCode】#290. Word Pattern

一天一道LeetCode

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（一）題目

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.
Examples:

• pattern = “abba”, str = “dog cat cat dog” should return true.
• pattern = “abba”, str = “dog cat cat fish” should return false.
• pattern = “aaaa”, str = “dog cat cat dog” should return false.
• pattern = “abba”, str = “dog dog dog dog” should return false.

Notes:

• You may assume pattern contains only lowercase letters, and str contains lowercase letters separated by a single space.

（二）解題

題目大意：字符串模式匹配。給定模板字符串pattern和待匹配字符串str，判斷str是否與pattern模式匹配
解題思路：這是一道典型應用hash表求解的題。
利用兩個輔助的hash表實現字符串的雙向匹配。例如a->dog，dog->a
不允許出現模式中兩個不同的字符對應str中同一個單詞。
具體思路見代碼：

class Solution {
public:
    bool wordPattern(string pattern, string str) {
        int lenp=pattern.length();
        int lens=str.length();
        int i = 0;
        int j = 0;
        unordered_map<string,char> hash;//兩個hash表，雙向匹配
        unordered_map<char,string> hash2;
        while(i<lenp&&j<lens){
            string temp;
            while(j<lens&&str[j]!=' '){//找出str中以空格分隔的字符串
                temp+=str[j++];
            }
            auto iter = hash.find(temp);
            auto iter2 = hash2.find(pattern[i]);
            if(iter==hash.end()&&iter2==hash2.end()){//如果都不存在
                hash[temp] = pattern[i];
                hash2[pattern[i]] = temp;//記錄雙向匹配關係
            }
            else{
                if(hash[temp]==pattern[i]&&hash2[pattern[i]]==temp) {}
                else return false;//匹配不上
            }
            i++;j++;
        }
        return i>=lenp?j>=lens?true:false:false;//最後需要判斷兩個字符串是否匹配完
    }
};