Leetcode 290. Word Pattern

最新推荐文章于 2024-01-10 12:35:36 发布

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本文介绍了一种模式匹配算法，该算法通过将输入字符串按空格分割成单词，并使用哈希表来建立单词与模式字符之间的对应关系，以此来判断字符串是否遵循给定的模式。文章提供了详细的实现步骤和示例代码。

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原题：

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.

解决方法：

- 先将字符串分成单词；

- 建立单词与模式字符之间的对应哈希表；

- 用哈希表判断是否符合规范。

代码：

std::vector<std::string> split(std::string str, std::string pattern)
	{
		std::string::size_type pos;
		std::vector<std::string> result;
		str += pattern;//扩展字符串以方便操作
		int size = str.size();

		for (int i = 0; i<size; i++)
		{
			pos = str.find(pattern, i);
			if (pos<size)
			{
				std::string s = str.substr(i, pos - i);
				result.push_back(s);
				i = pos + pattern.size() - 1;
			}
		}
		return result;
	}
    
    bool wordPattern(string pattern, string str) {
        std::vector<std::string> words = split(str, " ");
		if (words.size() != pattern.size())
			return false;
        
        map<char, string> mp1;
        map<string, char> mp2;
        for(int i = 0; i < words.size(); i++){
            if (!mp1.count(pattern[i]) && !mp2.count(words[i])){
                mp1[pattern[i]] = words[i];
                mp2[words[i]] = pattern[i];
            }
            else if (mp1.count(pattern[i]) && mp2.count(words[i])){
                if (mp1[pattern[i]] != words[i])
                    return false;
                
                if (mp2[words[i]] != pattern[i])
                    return false;
            }else{
                return false;
            }
        }
        return true;
    }