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.

注意这是一一映射，也就是说如果a->dog, b->dog，就应该return false，所以应该在HashMap基础上再加一层检查，即若不含该key，加入map之前应该检查map.values().contains(String)

第二遍：use HashMap和HashSet

public class Solution {
    public boolean wordPattern(String pattern, String str) {
        if (pattern==null || str==null) return false;
        String[] all = str.split(" ");
        
        if (pattern.length() != all.length) return false;
        HashMap<Character, String> map = new HashMap<Character, String>();
        HashSet<String> set = new HashSet<String>();
        for (int i=0; i<pattern.length(); i++) {
            char cur = pattern.charAt(i);
            if (!map.containsKey(cur)) {
                if (set.contains(all[i])) return false;
                map.put(pattern.charAt(i), all[i]);
                set.add(all[i]);
            }
            else {
                if (!map.get(cur).equals(all[i]))
                    return false;
            }
        }
        return true;
    }
}

　　用 map.values().contains()

public class Solution {
    public boolean wordPattern(String pattern, String str) {
        if (pattern==null || str==null) return false;
        String[] all = str.split(" ");
        
        if (pattern.length() != all.length) return false;
        HashMap<Character, String> map = new HashMap<Character, String>();
        for (int i=0; i<pattern.length(); i++) {
            if (!map.containsKey(pattern.charAt(i))) {
                if (map.values().contains(all[i])) return false;
                map.put(pattern.charAt(i), all[i]);
            }
            else {
                if (!map.get(pattern.charAt(i)).equals(all[i]))
                    return false;
            }
        }
        return true;
    }
}