212. Word Search II

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就是dfs+trie

1.先建立一个trie,把字典里所有的词加到trie里

2.对于板子里面的每个格子开始,向四个方向搜索,每次到了一个新的格子,添加在之前的单词上,然后检查trie,如果没有以这个开头的词,那就返回,如果包含了这个词,就加到结果里

要注意的是,即使包含这个词,还是要继续往下走,比如“aa”和“aab”都在词典中,不可以找到“aa”就停止了

 

  1 public class Solution {
  2     static final int[][] dirs = {{-1, 0}, {1, 0}, {0, 1}, {0, -1}};
  3     
  4     public List<String> findWords(char[][] board, String[] words) {
  5         Set<String> res = new HashSet<String>();
  6         Trie trie = new Trie();
  7         boolean[][] visited = new boolean[board.length][board[0].length];
  8         for(int i = 0; i < words.length; i++) {
  9             trie.insert(words[i]);
 10         }
 11         for(int i = 0; i < board.length; i++) {
 12             for(int j = 0; j < board[0].length; j++) {
 13                 dfs(res, board, "", i, j, trie, visited);
 14             }
 15         }
 16         return new ArrayList<String>(res);
 17     }
 18     
 19     private void dfs(Set<String> res, char[][] board, String word, int x, int y, Trie trie, boolean[][] visited) {
 20         if(x < 0 || y < 0 || x >= board.length || y >= board[0].length || visited[x][y]) {
 21             return;
 22         }
 23         word += board[x][y];
 24         if(!trie.startsWith(word)) {
 25             return;
 26         }
 27         if(trie.search(word)) {
 28             res.add(word);
 29         }
 30         visited[x][y] = true;
 31         for(int[] dir: dirs) {
 32             dfs(res, board, word, x + dir[0], y + dir[1], trie, visited);
 33         }
 34         visited[x][y] = false;
 35     }
 36     
 37     class TrieNode {
 38         char c;
 39         boolean isLeaf;
 40         TrieNode[] next;
 41         // Initialize your data structure here.
 42         public TrieNode() {
 43             next = new TrieNode[26];
 44             isLeaf = false;
 45         }
 46     }
 47     
 48     public class Trie {
 49         private TrieNode root;
 50     
 51         public Trie() {
 52             root = new TrieNode();
 53             
 54         }
 55     
 56         // Inserts a word into the trie.
 57         public void insert(String word) {
 58             if(word == null || word.length() == 0) {
 59                 return;
 60             }
 61             TrieNode cur = root;
 62             for(int i = 0; i < word.length(); i++) {
 63                 char c = word.charAt(i);
 64                 int index = c - ‘a‘;
 65                 if(cur.next[index] == null) {
 66                     cur.next[index] = new TrieNode();
 67                     cur.next[index].c = c;
 68                 }
 69                 cur = cur.next[index];
 70             }
 71             cur.isLeaf = true;
 72         }
 73     
 74         // Returns if the word is in the trie.
 75         public boolean search(String word) {
 76             if(word == null || word.length() == 0) {
 77                 return true;
 78             }
 79             TrieNode cur = root;
 80             for(int i = 0; i < word.length(); i++) {
 81                 char c = word.charAt(i);
 82                 int index = c - ‘a‘;
 83                 if(cur.next[index] == null) {
 84                     return false;
 85                 }
 86                 cur = cur.next[index];
 87             }
 88             return cur.isLeaf;
 89         }
 90     
 91         // Returns if there is any word in the trie
 92         // that starts with the given prefix.
 93         public boolean startsWith(String prefix) {
 94             if(prefix == null || prefix.length() == 0) {
 95                 return true;
 96             }
 97             TrieNode cur = root;
 98             for(int i = 0; i < prefix.length(); i++) {
 99                 char c = prefix.charAt(i);
100                 int index = c - ‘a‘;
101                 if(cur.next[index] == null) {
102                     return false;
103                 }
104                 cur = cur.next[index];
105             }
106             return true;
107         }
108     }
109             
110 }

 

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算法: TireNode解决搜索词212. Word Search II