时间轴

2025-11-28

init

题目：

P212 单词搜索 II

https://leetcode.cn/problems/word-search-ii/description/?envType=study-plan-v2&envId=top-interview-150

在 Trie 中保存字符串，这样避免递归时传递每次遍历的字符串。
用非递归的 DFS 版本对 board 的处理会有问题，DFS 递归刚好符合回溯。

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#include <vector>
#include <string>
#include <unordered_map>
#include <unordered_set>

using std::string;
using std::vector;
using std::unordered_map;
using std::unordered_set;

struct TrieNode {
	string word;
	unordered_map<char, TrieNode *> children;
	TrieNode()
	{
		this->word = "";
	}
};

void insertTrie(TrieNode *root, const string &word)
{
	TrieNode *node = root;
	for (auto c : word) {
		if (!node->children.count(c)) {
			node->children[c] = new TrieNode();
		}
		node = node->children[c];
	}
	node->word = word;
}

class Solution {
    public:
	int dirs[4][2] = { { 1, 0 }, { -1, 0 }, { 0, 1 }, { 0, -1 } };

	bool dfs(vector<vector<char> > &board, int x, int y, TrieNode *root, unordered_set<string> &res)
	{
		char ch = board[x][y];
		if (!root->children.count(ch)) {
			return false;
		}
		root = root->children[ch];
		if (root->word.size() > 0) {
			res.insert(root->word);
		}

		board[x][y] = '#';
		for (int i = 0; i < 4; ++i) {
			int nx = x + dirs[i][0];
			int ny = y + dirs[i][1];
			if (nx >= 0 && nx < board.size() && ny >= 0 && ny < board[0].size()) {
				if (board[nx][ny] != '#') {
					dfs(board, nx, ny, root, res);
				}
			}
		}
		board[x][y] = ch;

		return true;
	}

	vector<string> findWords(vector<vector<char> > &board, vector<string> &words)
	{
		TrieNode *root = new TrieNode();
		unordered_set<string> res;
		vector<string> ans;

		for (auto &word : words) {
			insertTrie(root, word);
		}
		for (int i = 0; i < board.size(); ++i) {
			for (int j = 0; j < board[0].size(); ++j) {
				dfs(board, i, j, root, res);
			}
		}

		ans.assign(res.begin(), res.end());
		return ans;
	}
};