题目1
给定一个 m x n 二维字符网格 board 和一个字符串单词 word 。如果 word 存在于网格中，返回 true ；否则，返回 false 。
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不允许被重复使用。
示例 1：
输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCCED" 输出：true
示例 2：
输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "SEE" 输出：true
示例 3：
输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCB" 输出：false
提示：
-
m == board.length
-
n = board[i].length
-
1 <= m, n <= 6
-
1 <= word.length <= 15
-
board 和 word 仅由大小写英文字母组成
参考答案
class Solution {
public:
bool check(vector<vector<char>>& board, vector<vector<int>>& visited, int i, int j, string& s, int k) {
if (board[i][j] != s[k]) {
return false;
} else if (k == s.length() - 1) {
return true;
}
visited[i][j] = true;
vector<pair<int, int>> directions{
{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
bool result = false;
for (const auto& dir: directions) {
int newi = i + dir.first, newj = j + dir.second;
if (newi >= 0 && newi < board.size() && newj >= 0 && newj < board[0].size()) {
if (!visited[newi][newj]) {
bool flag = check(board, visited, newi, newj, s, k + 1);
if (flag) {
result = true;
break;
}
}
}
}
visited[i][j] = false;
return result;
}
bool exist(vector<vector<char>>& board, string word) {
int h = board.size(), w = board[0].size();
vector<vector<int>> visited(h, vector<int>(w));
for (int i = 0; i < h; i++) {
for (int j = 0; j < w; j++) {
bool flag = check(board, visited, i, j, word, 0);
if (flag) {
return true;
}
}
}
return false;
}
};
题目2
定一个 m x n 二维字符网格 board 和一个单词（字符串）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。
示例 1：
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"]
示例 2：
输入：board = [["a","b"],["c","d"]], words = ["abcb"] 输出：[]
提示：
-
m == board.length
-
n == board[i].length
-
1 <= m, n <= 12
-
boardi 是一个小写英文字母
-
1 <= words.length <= 3 * 104
-
1 <= words[i].length <= 10
-
words[i] 由小写英文字母组成
-
words 中的所有字符串互不相同
参考答案
class TrieNode{
public:
string word = "";
vector<TrieNode*> nodes;
TrieNode():nodes(26, 0){}
};
class Solution {
int rows, cols;
vector<string> res;
public:
vector<string> findWords(vector<vector<char>>& board, vector<string>& words) {
rows = board.size();
cols = rows ? board[0].size():0;
if(rows==0 || cols==0) return res;
//建立字典树的模板
TrieNode* root = new TrieNode();
for(string word:words){
TrieNode *cur = root;
for(int i=0; i<word.size(); ++i){
int idx = word[i]-'a';
if(cur->nodes[idx]==0) cur->nodes[idx] = new TrieNode();
cur = cur->nodes[idx];
}
cur->word = word;
}
//DFS模板
for(int i=0; i<rows; ++i){
for(int j=0; j<cols; ++j){
dfs(board, root, i, j);
}
}
return res;
}
void dfs(vector<vector<char>>& board, TrieNode* root, int x, int y){
char c = board[x][y];
//递归边界
if(c=='.' || root->nodes[c-'a']==0) return;
root = root->nodes[c-'a'];
if(root->word!=""){
res.push_back(root->word);
root->word = "";
}
board[x][y] = '.';
if(x>0) dfs(board, root, x-1, y);
if(y>0) dfs(board, root, x, y-1);
if(x+1<rows) dfs(board, root, x+1, y);
if(y+1<cols) dfs(board, root, x, y+1);
board[x][y] = c;
}
};