LeetCode-212. Word Search II

LeetCode-212. 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.

Example:

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"]​​

​​题解：​​

​​使用字典树存储所有单词，节省存储空间。​​

字典树的路径若为单词，则标记为单词，访问后置否去重。

class Solution {public: struct Trie { vector child; bool isWord; Trie(): child(vector(26, NULL)), isWord(false){} }; Trie* buildTree(vector &words) { Trie *root = new Trie(); for (int i = 0; i < words.size(); i++) { Trie *t = root; for (int j = 0; j < words[i].size(); j++) { if (t->child[words[i][j] - 'a'] == NULL) { t->child[words[i][j] - 'a'] = new Trie(); } t = t->child[words[i][j] - 'a']; } t->isWord = true; } return root; } void dfs(vector &res, vector>& board, int n, int m, int x, int y, Trie *t, string &s) { if (x < 0 || x >= n || y < 0 || y >= m || t == NULL || board[x][y] == 'O') { return; } if (t->child[board[x][y] - 'a'] == NULL) { return; } s += board[x][y]; t = t->child[board[x][y] - 'a']; if (t->isWord == true) { res.push_back(s); t->isWord = false; } char ch = board[x][y]; board[x][y] = 'O'; dfs(res, board, n, m, x + 1, y, t, s); dfs(res, board, n, m, x - 1, y, t, s); dfs(res, board, n, m, x, y + 1, t, s); dfs(res, board, n, m, x, y - 1, t, s); board[x][y] = ch; s.pop_back(); } vector findWords(vector>& board, vector& words) { Trie *t = buildTree(words); if (board.empty() == true || words.empty() == true) { return {}; } int n = board.size(), m = board[0].size(); vector res; string s; for (int i = 0; i < n; i++) { for (int j = 0; j < m; j++) { dfs(res, board, n, m, i, j, t, s); } } return res; }};

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