[LeetCode] 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:
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 msclass 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 mspublic 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;    }}