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  • 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<Character, String> map = new HashMap<Character, String>();
     8         HashSet<String> set = new HashSet<String>();
     9         for (int i=0; i<pattern.length(); 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<Character, String> map = new HashMap<Character, String>();
     8         for (int i=0; i<pattern.length(); 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 }
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  • 原文地址:https://www.cnblogs.com/EdwardLiu/p/5081333.html
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