add solution of problem 290: word pattern

Author: ppulse

WordPattern290/README.md

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+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.

WordPattern290/Solution.java

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+public class Solution {
+    public boolean wordPattern(String pattern, String str) {
+        if (pattern == null || str == null)
+            return false;
+            
+        String[] strs = str.split(" ");
+        
+        if (pattern.length() != strs.length) {
+            return false;
+        } else {
+            Map<String, Character> map1 = new HashMap<String, Character>();
+            Map<Character, String> map2 = new HashMap<Character, String>();
+
+            for (int i = 0; i < pattern.length(); i++) {
+                if (!map1.containsKey(strs[i]) && !map2.containsKey(pattern.charAt(i))) {
+                    map1.put(strs[i], pattern.charAt(i));
+                    map2.put(pattern.charAt(i), strs[i]);
+                } else if (!map1.containsKey(strs[i]) || map1.get(strs[i]) != pattern.charAt(i)) {
+                    return false;
+                } else if (!map2.containsKey(pattern.charAt(i)) || !strs[i].equals(map2.get(pattern.charAt(i)))) {
+                    return false;
+                }
+            }
+            
+            return true;
+        }
+    }
+}