给定一个二维网格 board 和一个字典中的单词列表 words,找出所有同时在二维网格和字典中出现的单词。
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。
输入:
words = ["oath","pea","eat","rain"] and board =
[
['o','a','a','n'],
['e','t','a','e'],
['i','h','k','r'],
['i','f','l','v']
]
输出: ["eat","oath"]
Trie
class Trie {
struct TrieNode {
bool end;
string str;
unordered_map<char, TrieNode *> child;
TrieNode(): end(false), str("") {};
};
public:
Trie() {
root = new TrieNode();
}
void insert(string word) {
TrieNode *cur = root;
for (int i = 0; i < word.size(); i++) {
if (cur->child.count(word[i]) == 0) {
cur->child[word[i]] = new TrieNode();
}
cur = cur->child[word[i]];
}
cur->str = word; //下一个节点存单词, 当前节点不存。
cur->end = true;
}
void search(vector<string>& res, vector<vector<char>>& board) {
for (int i = 0; i < board.size(); i++) {
for (int j = 0; j < board[i].size(); j++) {
help(res, board, root, i, j);
}
}
}
void help(vector<string>& res, vector<vector<char>>& board, TrieNode *p, int x, int y) {
if (p->end) { //因为trie树把单词存入下一个节点,所以这里进来就判断是不是单词结尾,走到该处,说明有一个完整的单词
res.push_back(p->str);
p->end = false; //防止重复添加
return;
}
if (x < 0 || x == board.size() || y < 0 || y == board[x].size()) return; //判断边界
if (p->child.find(board[x][y]) == p->child.end()) return; //当前节点的map中找不到board中四通图的字符则返回
//下面是找到了字符
p = p->child[board[x][y]]; //找下一个节点,下一个节点end为true的话,说明找到了完整的单词
char cur = board[x][y];
board[x][y] = '#';
help(res, board, p, x + 1, y); //不能先算减一
help(res, board, p, x - 1, y);
help(res, board, p, x, y + 1); //同上
help(res, board, p, x, y - 1);
board[x][y] = cur;
}
private:
TrieNode *root;
};
class Solution {
public:
vector<string> findWords(vector<vector<char>>& board, vector<string>& words) {
Trie trie;
for (string &w : words) {
trie.insert(w);
}
vector<string> res;
trie.search(res, board);
return res;
}
};