[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：对这题，我也是感觉哔了狗了，无限WA，唉！本来想做几个水题的，又被虐了一顿。这道题是要求一一对应的关系，也就是说一个pattern对应一个str的子串，同时一个str的子串也要对应pattern的一个字母

代码如下：

class Solution {
public:
    bool wordPattern(string pattern, string str) {
        unordered_map<char, string> hash;
        unordered_map<string, char> hashR;
        vector<string> vec;
        int i = 0;
        while(i < str.size())//分割字符串，蛋疼啊，为何c++标准为什么还没有提供函数分割，连java都有^.^
        {
            size_t pos = str.find(' ', i);
            vec.push_back(str.substr(i, pos - i));
            if(pos == string::npos) break;
            i = pos+1;
        }
        if(pattern.size() != vec.size())//如果模型和字符串个数不一样，则不匹配
            return false;
        for(int i = 0; i< pattern.size(); i++)
        {
            if((hash.find(pattern[i]) == hash.end() && hashR.find(vec[i]) == hashR.end())//如果原来不存在
                || (hash[pattern[i]] == vec[i] && hashR[vec[i]] == pattern[i]))//原来存在，并且和现在对应关系相同
            {
                hash[pattern[i]] = vec[i];
                hashR[vec[i]] = pattern[i];
            }
            else
                return false;
        }
        return true;
    }
};