問題描述：

Given apatternand a stringstr, find ifstrfollows the same pattern.

Herefollowmeans a full match, such that there is a bijection between a letter inpatternand anon-emptyword instr.

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.

利用hashtable，將pattern中的每個字元對映成int，str的每個單詞同樣也對映成int，兩者對比，若有對映不相同的，即視為不匹配。

過程詳見程式碼：

class Solution {
public:
    bool wordPattern(string pattern, string str) {
        map<char,int> p2i;
        map<string,int> w2i;
        istringstream in(str);
        int i = 0, n = pattern.size();
        for(string word; in >> word; i++)
        {
        	if(i == n || p2i[pattern[i]] != w2i[word]) return false;
        	p2i[pattern[i]] = w2i[word] = i + 1;
		}
		return i == n;
    }
};