Given an m x nboard of characters and a list of strings words, return all words on the board.
Each word must be constructed from letters of sequentially adjacent cells, where adjacent cells are horizontally or vertically neighboring. The same letter cell may not be used more than once in a word.
Example 1:
**Input:** board = [["o","a","a","n"],["e","t","a","e"],["i","h","k","r"],["i","f","l","v"]], words = ["oath","pea","eat","rain"]
**Output:** ["eat","oath"]
Example 2:
**Input:** board = [["a","b"],["c","d"]], words = ["abcb"]
**Output:** []
Constraints:
m == board.length
n == board[i].length
1 <= m, n <= 12
board[i][j] is a lowercase English letter.
1 <= words.length <= 3 * 104
1 <= words[i].length <= 10
words[i] consists of lowercase English letters.
All the strings of words are unique.
ac
class Solution {
public List<String> findWords(char[][] board, String[] words) {
List<String> res = new ArrayList<String>();
// edge cases
if (board.length == 0 || board[0].length == 0 || words.length == 0) {
return res;
}
// build trie
TrieNode root = new TrieNode();
root.build(words);
for (int r = 0; r < board.length; r++) {
for (int c = 0; c < board[0].length; c++) {
dfs(board, r, c, root, res);
}
}
return res;
}
private void dfs(char[][] board, int r, int c, TrieNode node, List<String> res) {
// out of boundary?
if (r < 0 || c < 0 || r >= board.length || c >= board[0].length) return;
// visited in this word? current char not in trie? return
int pos = board[r][c] - 'a';
if (board[r][c] == '#' || node.next[pos] == null) return;
// continue, mark visited
char curr = board[r][c];
board[r][c] = '#';
// find word
if (node.next[pos].word != null) {
res.add(node.next[pos].word);
node.next[pos].word = null;
}
// adjacent chars
dfs(board, r+1, c, node.next[pos], res);
dfs(board, r-1, c, node.next[pos], res);
dfs(board, r, c+1, node.next[pos], res);
dfs(board, r, c-1, node.next[pos], res);
// resume visited mark
board[r][c] = curr;
}
}
class TrieNode {
public TrieNode[] next;
String word;
public TrieNode() {
next = new TrieNode[26];
}
public void insert(String s, int index) {
if (index == s.length()) {
this.word = s;
return;
}
int i = s.charAt(index) - 'a';
if (next[i] == null) next[i] = new TrieNode();
next[i].insert(s, index+1);
}
public void build(String[] words){
for (String w : words) {
insert(w, 0);
}
}
}
// build a trie, node with a string
// walk board, current char in trie? continue : return;
//board[r][c] - 'a'
