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.

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

 1 public class Solution {
 2     public boolean wordPattern(String pattern, String str) {
 3         if (pattern==null || str==null) return false;
 4         String[] all = str.split(" ");
 5         
 6         if (pattern.length() != all.length) return false;
 7         HashMap map = new HashMap();
 8         HashSet set = new HashSet();
 9         for (int i=0; i) {
10             char cur = pattern.charAt(i);
11             if (!map.containsKey(cur)) {
12                 if (set.contains(all[i])) return false;
13                 map.put(pattern.charAt(i), all[i]);
14                 set.add(all[i]);
15             }
16             else {
17                 if (!map.get(cur).equals(all[i]))
18                     return false;
19             }
20         }
21         return true;
22     }
23 }

第一遍：

 1 public class Solution {
 2     public boolean wordPattern(String pattern, String str) {
 3         if (pattern==null || str==null) return false;
 4         String[] all = str.split(" ");
 5         
 6         if (pattern.length() != all.length) return false;
 7         HashMap map = new HashMap();
 8         for (int i=0; i) {
 9             if (!map.containsKey(pattern.charAt(i))) {
10                 if (map.values().contains(all[i])) return false;
11                 map.put(pattern.charAt(i), all[i]);
12             }
13             else {
14                 if (!map.get(pattern.charAt(i)).equals(all[i]))
15                     return false;
16             }
17         }
18         return true;
19     }
20 }