题意
题解
DFS+回溯
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;
}
};