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:
pattern = "abba", str = "dog cat cat dog" should return true.
pattern = "abba", str = "dog cat cat fish" should return false.
pattern = "aaaa", str = "dog cat cat dog" should return false.
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.

解题思路

• 用一个哈希表建立从pattern中字符到str中字符串的映射。
• 用一个集合表示str中字符串是否已作为值出现在哈希表中。
  以下两种情况为假：
  
  • 哈希表中已存在key，但是对应的value值不一样。
  • 哈希表中不存在key，但是对应的值已作为其他key的value值。

实现代码

C++:

// Runtime: 0 ms
class Solution {
public:
    bool wordPattern(string pattern, string str) {
        vector<string> words;
        istringstream iss(str);
        string temp;
        while (iss>>temp)
        {
            words.push_back(temp);
        }
        if (words.size() != pattern.size())
        {
            return false;
        }
        unordered_map<char, string> mymap;
        set<string> used;
        for (int i = 0; i < pattern.length(); i++)
        {
            if (mymap.find(pattern[i]) == mymap.end())
            {
                if (used.find(words[i]) == used.end())
                {
                    mymap[pattern[i]] = words[i];
                    used.insert(words[i]);
                }
                else
                {
                    return false;
                }
            }
            else if (mymap[pattern[i]].compare(words[i]) != 0)
            {
                return false;
            }
        }

        return true;
    }
};

Java:

// Runtime: 3 ms
public class Solution {
    public boolean wordPattern(String pattern, String str) {
        String words[] = str.split(" ");
        if (pattern.length() != words.length) {
            return false;
        }
        Map<Character, String> mymap = new HashMap<Character, String>();
        for (int i = 0; i < pattern.length(); i++) {
            if (!mymap.containsKey(pattern.charAt(i))) {
                if (!mymap.containsValue(words[i])) {
                    mymap.put(pattern.charAt(i), words[i]);
                }
                else {
                    return false;
                }
            }
            else if (!mymap.get(pattern.charAt(i)).equals(words[i])) {
                return false;
            }
        }

        return true;
    }
}