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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 substring in str.
Examples:
"abab", str = "redblueredblue" should return true."aaaa", str = "asdasdasdasd" should return true."aabb", str = "xyzabcxzyabc" should return false.
Notes:
You may assume both pattern and str contains only lowercase letters.
Word Pattern 是用HashMap的containsKey()和containsValue()两个方法实现的。
但是对于Word Pattern II，我们并不知道怎样划分str，所以基本想法是“暴力搜索”即backtracking/DFS。这里DFS的存储对象不是常见的list，而是map.
1 public class Solution { 2 public boolean wordPatternMatch(String pattern, String str) { 3 return helper(pattern, 0, str, 0, new HashMap<Character, String>()); 4 } 5 6 private boolean helper(String pattern, int startP, String str, int startS, Map<Character, String> map) { 7 if (startP == pattern.length() && startS == str.length()) { 8 return true; 9 } else if (startP == pattern.length()) { 10 11 } 12 if (match.equals(str.substring(startS, endS))) { 13 return helper(pattern, startP + 1, str, endS, map); 14 } else { 15 return false; 16 } 17 } else { 18 // If map does not have existing key 19 // Traverse, brute force 20 21 for (int i = startS + 1; i <= str.length(); i++) { 22 String candidate = str.substring(startS, i); 23 if (map.containsValue(candidate)) { 24 continue; 25 } 26 map.put(key, candidate); 27 if (helper(pattern, startP + 1, str, i, map)) { 28 return true; 29 } 30 map.remove(key); 31 } 32 } 33 return false; 34 } 35 }
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原文地址：http://www.cnblogs.com/ireneyanglan/p/4946364.html
