2684. 矩阵中移动的最大次数 【BFS， DFS 记得标记访问过的位置】

2684. 矩阵中移动的最大次数

class Solution:
    def maxMoves(self, grid: List[List[int]]) -> int: 
        ret = 0  # 次数
        m, n = len(grid), len(grid[0])
        # 注意 只能 从 第 1 列 的任一格 出发 
        visited = set()                
        for i in range(m):
            if (i, 0) in visited: continue
            q = collections.deque([(i, 0, 0)])
            visited.add((i, 0))
            while q:
                x, y, step = q.popleft()
                ret = max(ret, step)
                for nx, ny in (x - 1, y + 1), (x, y + 1), (x + 1, y + 1):
                    if 0 <= nx < m and 0 <= ny < n and grid[nx][ny] > grid[x][y] and (nx, ny) not in visited:
                        q.append((nx, ny, step + 1))
                        visited.add((nx, ny))

        return ret

class Solution:
    def maxMoves(self, grid: List[List[int]]) -> int:
        self.ret = 0
        m, n = len(grid), len(grid[0])
        visited = set()
        def dfs(x, y, step):
            self.ret = max(self.ret, step)   
            visited.add((x, y))         
            for nx, ny in (x - 1, y + 1), (x, y + 1), (x + 1, y + 1):
                if 0 <= nx < m and 0 <= ny < n and grid[nx][ny] > grid[x][y] and (nx, ny) not in visited:
                    dfs(nx, ny, step + 1)

        for i in range(m):
            dfs(i, 0, 0)

        return self.ret

unordered_set文档

class Solution {
public:
    int maxMoves(vector<vector<int>>& grid) {
        int m = grid.size(), n = grid[0].size();
        int ret = 0;  //  最大步数
        vector<vector<int>> visited(m, vector<int>(n));  // 标记 是否 访问过
        for (int i = 0; i < m; ++i){
            if (visited[i][0])  continue; // 已访问 跳过
            queue<vector<int>> q;
            q.push({i, 0, 0});
            visited[i][0] = true;
            while (!q.empty()){
                int x = q.front()[0], y = q.front()[1], step = q.front()[2]; q.pop();
                ret = max(ret, step);
                int dx[3] = {-1, 0, 1};
                int dy[3] = {1, 1, 1};
                for (int idx = 0; idx < 3; ++idx){
                    int nx = x + dx[idx], ny = y + dy[idx];
                    if (nx >= 0 && nx < m && ny >= 0 && ny < n && grid[nx][ny] > grid[x][y] && !visited[nx][ny]){
                        q.push({nx, ny, step + 1});
                        visited[nx][ny] = true; // 标记 已访问
                    }
                }
                
            }
        }
        return ret;       
        
    }
};