LeetCode 1091 Shortest Path in Binary Matrix

Given an n x n binary matrix grid, return the length of the shortest clear path in the matrix. If there is no clear path, return -1.

A clear path in a binary matrix is a path from the top-left cell (i.e., (0, 0)) to the bottom-right cell (i.e., (n - 1, n - 1)) such that:

• All the visited cells of the path are 0.
• All the adjacent cells of the path are 8-directionally connected (i.e., they are different and they share an edge or a corner).

The length of a clear path is the number of visited cells of this path.

Solution

利用 $BFS$ 即可。只不过这次的 $dir$ 有 $8$ 个方向。另外一点就是，我们可以直接在 $grid$ 上面进行修改：修改为路径长度，当然判断的条件为：

• 没有被访问： $grid[i][j]=0$
• 符合边界条件

点击查看代码

class Solution {
private:
    int ans = -1;
    queue<vector<int>> q;
    
    int dir[8][2] = {
        -1,0,
        0,-1,
        1,0,
        0,1,
        -1,-1,
        1,-1,
        1,1,
        -1,1
    };
    
    bool check(int i,int j,int r, int c){
        if(i<0||j<0||i>=r||j>=c)return false;
        return  true;
    }
    
    
    
public:
    int shortestPathBinaryMatrix(vector<vector<int>>& grid) {
        int r = grid.size(), c = grid[0].size();
        if(grid[0][0])return -1;
        grid[0][0]=1;
        q.push({0,0});
        while(!q.empty()){
            auto f = q.front();q.pop();
            int x = f[0], y = f[1];
            if(x==r-1 && y==c-1)return grid[x][y];
            for(int i=0;i<8;i++){
                int nx = x+dir[i][0], ny = y+dir[i][1];
                if(check(nx,ny,r,c) && grid[nx][ny]==0){
                    grid[nx][ny] = grid[x][y]+1;
                    q.push({nx,ny});
                }
            }
        }
        return -1;
    }
};