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.

1.我的答案

這裡有個技巧：就是對兩個字串，記錄每個字元當前位置，當前比較的字元的之前位置相同，則將當前位置記錄；若比較的字元的之前位置不同，則不符合模式；若之前沒出現過，則將該字元的位置記為當前位置。

類似的題目為：205. Isomorphic Strings

以下為我的程式碼：

class Solution {
public:
    bool wordPattern(string pattern, string str) {
        map<char,int>map_p;
        map<string,int>map_s;
        vector<string> vec;
        string s = "";
        int len_pat = pattern.size();
        for(int i = 0; i < str.size(); i++){
            if(str[i] == ' '){
                vec.push_back(s);
                s = "";
                continue;
            }
            s += str[i];
        }
        vec.push_back(s);
        int len_str = vec.size();
        if(len_pat != len_str)
        return false;
        for(int k = 0; k < len_str; k++){
            if(map_p.find(pattern[k]) == map_p.end())
                map_p[pattern[k]] = k;
            if(map_s.find(vec[k]) == map_s.end())
                map_s[vec[k]] = k;
            if(map_p[pattern[k]] == map_s[vec[k]]){
                map_p[pattern[k]] = k;
                map_s[vec[k]] = k;
            }else
            return false;
        }
        return true;
    }
};
 

用i+1避免掉0產生的歧義性（雖然我現在還是不明白為什麼要這麼做）

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;
}