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.

Credits:
Special thanks to @minglotus6 for adding this problem and creating all test cases.

我开始以为模式都是aabb这种，就是都是a开始这种。不会出现eebb这样的，后来基于这种错误的改造，得到以下方法。

public boolean wordPattern(String pattern, String str) {String[] strs = str.split(" ");if (pattern.length() != strs.length)return false;Map<String, Character> map = new HashMap<String, Character>();String temp = "";char c = 'a';for (int i = 0; i < strs.length; i++) {if (!map.containsKey(strs[i]))map.put(strs[i], c++);temp = temp + map.get(strs[i]);}return temp.equals(transformPattern(pattern));}public String transformPattern(String s) {char c = 'a';String temp = "";char[] chs = s.toCharArray();Map<Character, Character> map = new HashMap<Character, Character>();for (int i = 0; i < s.length(); i++) {if (!map.containsKey(chs[i]))map.put(chs[i], c++);temp = temp + map.get(chs[i]);}return temp;}

如果再弄个set的话，一次遍历就能完成。

public boolean wordPattern(String pattern, String str) {String[] strs = str.split(" ");if (pattern.length() != strs.length)return false;Map<Character, String> map = new HashMap<Character, String>();Set<String> unique = new HashSet<String>();for (int i = 0; i < pattern.length(); i++) {char c = pattern.charAt(i);if (map.containsKey(c)) {if (!map.get(c).equals(strs[i]))return false;} else {if (unique.contains(strs[i]))return false;map.put(c, strs[i]);unique.add(strs[i]);}}return true;}