题目
给定一个 m x n 二维字符网格 board 和一个单词（字符串）列表 words， 返回所有二维网格上的单词 。
单词必须按照字母顺序，通过 相邻的单元格 内的字母构成，其中“相邻”单元格是那些水平相邻或垂直相邻的单元格。同一个单元格内的字母在一个单词中不允许被重复使用。
示例 1：
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| 输入：board = [["o","a","a","n"],["e","t","a","e"],["i","h","k","r"],["i","f","l","v"]], words = ["oath","pea","eat","rain"]
输出：["eat","oath"]
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示例 2：
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| 输入：board = [["a","b"],["c","d"]], words = ["abcb"]
输出：[]
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提示：
m == board.lengthn == board[i].length1 <= m, n <= 12board[i][j] 是一个小写英文字母1 <= words.length <= 3 * 1041 <= words[i].length <= 10words[i] 由小写英文字母组成words 中的所有字符串互不相同
题解
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| class Solution {
Set<String> set = new HashSet<>();
char[][] board;
int n, m;
TrieNode root = new TrieNode();
int[][] dirs = new int[][] { { 1, 0 }, { -1, 0 }, { 0, 1 }, { 0, -1 } };
boolean[][] vis = new boolean[15][15];
class TrieNode {
String s;
TrieNode[] tns = new TrieNode[26];
}
void insert(String s) {
TrieNode p = root;
for (int i = 0; i < s.length(); i++) {
int u = s.charAt(i) - 'a';
if (p.tns[u] == null)
p.tns[u] = new TrieNode();
p = p.tns[u];
}
p.s = s;
}
public List<String> findWords(char[][] _board, String[] words) {
board = _board;
m = board.length;
n = board[0].length;
for (String w : words)
insert(w);
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
int u = board[i][j] - 'a';
if (root.tns[u] != null) {
vis[i][j] = true;
dfs(i, j, root.tns[u]);
vis[i][j] = false;
}
}
}
List<String> ans = new ArrayList<>();
for (String s : set)
ans.add(s);
return ans;
}
void dfs(int i, int j, TrieNode node) {
if (node.s != null)
set.add(node.s);
for (int[] d : dirs) {
int dx = i + d[0], dy = j + d[1];
if (dx < 0 || dx >= m || dy < 0 || dy >= n)
continue;
if (vis[dx][dy])
continue;
int u = board[dx][dy] - 'a';
if (node.tns[u] != null) {
vis[dx][dy] = true;
dfs(dx, dy, node.tns[u]);
vis[dx][dy] = false;
}
}
}
}
|
