leetcode 212: Word Search II
Word Search II
Total Accepted: 423 Total Submissions: 2510Given a 2D board and a list of words from the dictionary, find all words in the board.
Each word must be constructed from letters of sequentially adjacent cell, where "adjacent" cells are those horizontally or vertically neighboring. The same letter cell may not be used more than once in a word.
For example,
Given words = ["oath","pea","eat","rain"] and board =
[ ['o','a','a','n'], ['e','t','a','e'], ['i','h','k','r'], ['i','f','l','v'] ]Return
["eat","oath"].
Note:
You may assume that all inputs are consist of lowercase letters a-z.
click to show hint.
You would need to optimize your backtracking to pass the larger test. Could you stop backtracking earlier?
If the current candidate does not exist in all words' prefix, you could stop backtracking immediately. What kind of data structure could answer such query efficiently? Does a hash table work? Why or why not? How about a Trie? If you would like to learn how to implement a basic trie, please work on this problem: Implement Trie (Prefix Tree) first.
[思路]
1. 按照word search I的思路, 超时.
2. trie 前缀树, 可以pruning 剪枝. pass.
[CODE]
1. 超时
public class Solution {
public List findWords(char[][] board, String[] words) {
List res = new ArrayList();
if(board==null || words==null || board.length==0 || words.length==0) return res;
boolean[][] visited = new boolean[board.length][board[0].length];
Set dict = new HashSet(Arrays.asList(words));
for(int i=0; i dict,int i,int j, StringBuilder sb, List res) {
if(i<0 || i>board.length-1 || j<0 || j>board[0].length-1 || visited[i][j]) return;
sb.append(board[i][j]);
visited[i][j] = true;
if(dict.contains(sb.toString()))
res.add(sb.toString());
search(board, visited, dict, i-1, j, sb, res);
search(board, visited, dict, i+1, j, sb, res);
search(board, visited, dict, i, j-1, sb, res);
search(board, visited, dict, i, j+1, sb, res);
sb.deleteCharAt(sb.length() - 1);
visited[i][j] = false;
}
} 2. 使用trie, 检查前缀, pruning
public class Solution {
public List findWords(char[][] board, String[] words) {
Set res = new HashSet();
if(board==null || words==null || board.length==0 || words.length==0) return new ArrayList(res);
boolean[][] visited = new boolean[board.length][board[0].length];
Trie trie = new Trie();
for(String word : words) {
trie.insert(word);
}
for(int i=0; i(res);
}
private void search(char[][] board,boolean[][] visited,Trie trie,int i,int j, StringBuilder sb, Set res) {
if(i<0 || i>board.length-1 || j<0 || j>board[0].length-1 || visited[i][j]) return;
sb.append(board[i][j]);
String s = sb.toString();
visited[i][j] = true;
if(trie.startsWith(s)) {
if(trie.search(s)) res.add(s);
search(board, visited, trie, i-1, j, sb, res);
search(board, visited, trie, i+1, j, sb, res);
search(board, visited, trie, i, j-1, sb, res);
search(board, visited, trie, i, j+1, sb, res);
}
sb.deleteCharAt(sb.length() - 1);
visited[i][j] = false;
}
}
class TrieNode {
// Initialize your data structure here.
char c;
boolean leaf;
HashMap children = new HashMap();
public TrieNode(char c) {
this.c = c;
}
public TrieNode(){};
}
class Trie {
private TrieNode root;
public Trie() {
root = new TrieNode();
}
// Inserts a word into the trie.
public void insert(String word) {
Map children = root.children;
for(int i=0; i children = root.children;
TrieNode t = null;
for(int i=0; i