290. Word Pattern-优快云博客

本文链接：https://blog.youkuaiyun.com/FreeeLinux/article/details/61784348

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，该算法利用unordered_map实现，确保模式字符串中的每个字符与目标字符串中的非空单词一一对应。通过具体实例展示了如何判断目标字符串是否遵循给定的模式。

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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.

使用unordered_map，时间复杂度O(n)：

class Solution {
public:
    bool wordPattern(string pattern, string str) {
        const int len1 = pattern.length();
        const int len2 = str.length();
        if(len1 == 0 && len2 == 0)
            return true;
        if(len1 == 0 || len2 == 0)
            return false;
        unordered_map<char, string> umap1;
        unordered_map<string, char> umap2;
        int j = -1, i = 0;
        for(i=0; i<len1; ++i){
            int c = j + 1;
            while(str[++j] != ' ' && str[j] != '\0');
            string tmp(str, c, j-c);
            auto it1 = umap1.insert(make_pair(pattern[i], tmp));
            auto it2 = umap2.insert(make_pair(tmp, pattern[i])); 
            if(!it1.second || !it2.second){
                if(it1.first->second != tmp || it2.first->second != pattern[i])
                    return false;
            }
            if(str[j] == '\0')  //注意为'\0'的情况
                break;
        }
        return j == str.length() && i == pattern.length()-1;  //注意两个匹配长度不能的情况
    }
};