题目：

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.

采取的办法是建立两个map，相互映射。为什么用一个map不够？如果面对第4个例子，abba，dog dog dog dog，a和b都会映射到dog，因为你要保证1对1映射，那么可以采取反面映射过来。

class Solution {public:    bool wordPattern(string pattern, string str) {        unordered_map
     
       dictP;        unordered_map
      
        dictS;        string buff;        int idx = 0;        for (int i = 0; i < pattern.size(); i++){            buff.clear();            while (idx < str.size() && str[idx] != ' '){                buff.push_back(str[idx++]);            }            if (buff.empty()) return false;            idx++;                        if (dictP.find(pattern[i]) == dictP.end() && dictS.find(buff) == dictS.end()){                dictP[pattern[i]] = buff;                dictS[buff] = pattern[i];            }            else if (dictP[pattern[i]] != buff || dictS[buff] != pattern[i])                 return false;        }        if (idx == 0 || idx < str.size())            return false;        return true;    }};