LeetCode 212. Word Search II（单词搜索）

原题网址：https://leetcode.com/problems/word-search-ii/

Given 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.

方法：前缀树+深度优先搜索。

public class Solution {
private TrieNode root = new TrieNode();
private int[] ro = {-1, 1, 0, 0};
private int[] co = {0, 0, -1, 1};
private void find(char[][] board, boolean[][] visited, int row, int col, TrieNode node, Set<String> founded) {
visited[row][col] = true;
TrieNode current = node.nexts[board[row][col]-'a'];
if (current.word != null) founded.add(current.word);
for(int i=0; i<4; i++) {
int nr = row + ro[i];
int nc = col + co[i];
if (nr < 0 || nr >= board.length || nc < 0 || nc >= board[nr].length || visited[nr][nc]) continue;
TrieNode next = current.nexts[board[nr][nc]-'a'];
if (next != null) find(board, visited, nr, nc, current, founded);
}
visited[row][col] = false;
}
public List<String> findWords(char[][] board, String[] words) {
Set<String> founded = new HashSet<>();
for(int i=0; i<words.length; i++) {
char[] wa = words[i].toCharArray();
TrieNode node = root;
for(int j=0; j<wa.length; j++) node = node.append(wa[j]);
node.word = words[i];
}
boolean[][] visited = new boolean[board.length][board[0].length];
for(int i=0; i<board.length; i++) {
for(int j=0; j<board[i].length; j++) {
if (root.nexts[board[i][j]-'a'] != null) find(board, visited, i, j, root, founded);
}
}
List<String> results = new ArrayList<>();
results.addAll(founded);
return results;
}
}
class TrieNode {
String word;
TrieNode[] nexts = new TrieNode[26];
TrieNode append(char ch) {
if (nexts[ch-'a'] != null) return nexts[ch-'a'];
nexts[ch-'a'] = new TrieNode();
return nexts[ch-'a'];
}
}

可以通过修改Trie数据的方式防止重复添加单词：

public class Solution {
private int[] dy = {-1, 1, 0, 0};
private int[] dx = {0, 0, -1, 1};
private void find(char[][] board, boolean[][] visit, int y, int x, Trie node, List<String> results) {
if (y < 0 || y >= board.length || x < 0 || x >= board[y].length) return;
if (visit[y][x]) return;
visit[y][x] = true;
Trie next = node.nexts[board[y][x]-'a'];
if (next != null) {
if (next.word != null) {
results.add(next.word);
next.word = null;
}
for(int i=0; i<4; i++) {
int ny = y+dy[i];
int nx = x+dx[i];
find(board, visit, ny, nx, next, results);
}
}
visit[y][x] = false;
}
public List<String> findWords(char[][] board, String[] words) {
Trie root = new Trie();
for(String word: words) {
Trie node = root;
char[] wa = word.toCharArray();
for(char c: wa) node = node.append(c);
node.word = word;
}
boolean[][] visit = new boolean[board.length][board[0].length];
List<String> results = new ArrayList<>();
for(int i=0; i<board.length; i++) {
for(int j=0; j<board[i].length; j++) {
find(board, visit, i, j, root, results);
}
}
return new ArrayList<>(results);
}
}
class Trie {
String word;
Trie[] nexts = new Trie[26];
Trie append(char ch) {
if (nexts[ch-'a'] == null) nexts[ch-'a'] = new Trie();
return nexts[ch-'a'];
}
}