[leetcode] 290. Word Pattern 解题报告

题目链接：https://leetcode.com/problems/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.

Examples:

1. pattern = "abba", str = "dog cat cat dog" should return true.
2. pattern = "abba", str = "dog cat cat fish" should return false.
3. pattern = "aaaa", str = "dog cat cat dog" should return false.
4. 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：一个字母对应一个字符串, 同时一个字符串也要对应一个字符, 这样我们可以同时用两个hash表来检测是否同时匹配. 还有需要注意的是模式串和匹配串长度不一致的问题.

代码如下：

class Solution {public:    bool wordPattern(string pattern, string str) {        str += " ";        int i = 0, j = 0, len1 = pattern.size(), len2=str.size();        unordered_map<char, string> hash1;        unordered_map<string, char> hash2;        while(i < len1 && j < len2)        {            int pos = str.find(' ', j);            string s = str.substr(j, pos-j);            if(hash1.count(pattern[i]) && hash1[pattern[i]]!=s) return false;            if(hash2.count(s) && hash2[s]!= pattern[i]) return false;            hash1[pattern[i]] = s, hash2[s] = pattern[i];            i++, j = pos+1;        }        return i==len1 && j ==len2;    }};

class Solution {public:    bool wordPattern(string pattern, string str) {        unordered_map<string, char> hash1;        unordered_map<char, string> hash2;        istringstream is(str);        for(auto ch: pattern)        {            if(is.eof()) return false;            string s;            is >> s;            if(!hash1.count(s) && !hash2.count(ch))            {                hash1[s] = ch, hash2[ch] = s;                continue;            }            if(hash1.count(s) && hash2.count(ch) && hash1[s]==ch && hash2[ch]==s)                 continue;            return false;        }        return is.eof();    }};