首页 > 代码库 > 212. Word Search II
212. Word Search II
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 }
212. Word Search II
声明:以上内容来自用户投稿及互联网公开渠道收集整理发布,本网站不拥有所有权,未作人工编辑处理,也不承担相关法律责任,若内容有误或涉及侵权可进行投诉: 投诉/举报 工作人员会在5个工作日内联系你,一经查实,本站将立刻删除涉嫌侵权内容。