290. Word Pattern

最新推荐文章于 2024-03-05 15:25:25 发布

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最新推荐文章于 2024-03-05 15:25:25 发布

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本文链接：https://blog.youkuaiyun.com/u013248806/article/details/69349183

本文介绍了一种基于模式字符串和目标字符串的匹配算法实现。通过使用双向哈希表确保模式与字符串间存在一一对应关系，实现了全匹配判断。文章详细解释了如何分割目标字符串并验证其与模式是否完全匹配。

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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.

【思路】 c++ 没有split函数所以需要手动分割str；第一次commit失败了发现 abba 与 “cat cat cat cat” 这个漏洞

所以要用两个hashtable 双向验证

    class Solution {  
    public:  
      bool wordPattern(string pattern, string str) {  
        unordered_map<char,string> map;  
        unordered_map<string,char> map2;  
        vector<string> vec;  
        int j=0;  
        for(int i=0;i<str.size();i++){  
            if(str[i]==' '){  
                string tmp =str.substr(j,i-j);  
                vec.push_back(tmp);  
                j=i+1;  
            }  
            if(i==str.size()-1){  
                string tmp =str.substr(j,i-j+1);  
                vec.push_back(tmp);  
            }  
      
        }  
        if(pattern.size()!=vec.size())return false;  
        for(int i=0;i<pattern.size();i++){  
            if(map.find(pattern[i])==map.end()&&map2.find(vec[i])==map2.end()){  
                map.insert(make_pair(pattern[i],vec[i]));  
                map2.insert(make_pair(vec[i],pattern[i]));  
            }  
            else if(map[pattern[i]]!=vec[i]||map2[vec[i]]!=pattern[i])return false;  
      
        }  
      
        return true;  
      
    }  
      
    };