980. Unique Paths III

On a 2-dimensional grid, there are 4 types of squares:

1 represents the starting square. There is exactly one starting square.
2 represents the ending square. There is exactly one ending square.
0 represents empty squares we can walk over.
-1 represents obstacles that we cannot walk over.

Return the number of 4-directional walks from the starting square to the ending square, that walk over every non-obstacle square exactly once.

Example 1:

Input: [[1,0,0,0],[0,0,0,0],[0,0,2,-1]]
Output: 2
Explanation: We have the following two paths: 
1. (0,0),(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(1,0),(2,0),(2,1),(2,2)
2. (0,0),(1,0),(2,0),(2,1),(1,1),(0,1),(0,2),(0,3),(1,3),(1,2),(2,2)

Example 2:

Input: [[1,0,0,0],[0,0,0,0],[0,0,0,2]]
Output: 4
Explanation: We have the following four paths: 
1. (0,0),(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(1,0),(2,0),(2,1),(2,2),(2,3)
2. (0,0),(0,1),(1,1),(1,0),(2,0),(2,1),(2,2),(1,2),(0,2),(0,3),(1,3),(2,3)
3. (0,0),(1,0),(2,0),(2,1),(2,2),(1,2),(1,1),(0,1),(0,2),(0,3),(1,3),(2,3)
4. (0,0),(1,0),(2,0),(2,1),(1,1),(0,1),(0,2),(0,3),(1,3),(1,2),(2,2),(2,3)

Example 3:

Input: [[0,1],[2,0]]
Output: 0
Explanation: 
There is no path that walks over every empty square exactly once.
Note that the starting and ending square can be anywhere in the grid.

Note:

1 <= grid.length * grid[0].length <= 20

class Solution {
    public int uniquePathsIII(int[][] grid) {
        int n = grid.length * grid[0].length;
        int[] res = new int[1];
        for(int i = 0; i < grid.length; i++) {
            for(int j = 0; j < grid[0].length; j++) {
                if(grid[i][j] == 1) dfs(grid, i, j, res, 0, n);
            }
        }
        return res[0];
    }
    public void dfs(int[][] grid, int i, int j, int[] res, int count, int n) {
        int cur = grid[i][j];
        if(cur == 2 && count == n - 1) {
            res[0]++;
            return;
        }
        if(cur == -1) return;
        
        grid[i][j] = -1;
        if(i > 0) dfs(grid, i - 1, j, res, count + 1, n);
        if(i < grid.length - 1) dfs(grid, i + 1, j, res, count + 1, n);
        if(j > 0) dfs(grid, i, j - 1, res, count + 1, n);
        if(j < grid[0].length - 1) dfs(grid, i, j + 1, res, count + 1, n);
        grid[i][j] = cur;
    }
}

38/39?尼玛

class Solution {
    int res = 0, empty = 1, sx, sy;
    public int uniquePathsIII(int[][] grid) {
        int m = grid.length, n = grid[0].length;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j] == 0) empty++;
                else if (grid[i][j] == 1) {
                    sx = i;
                    sy = j;
                }
            }
        }
        dfs(grid, sx, sy);
        return res;
    }

    public void dfs(int[][] grid, int x, int y) {
        if (x < 0 || x >= grid.length || y < 0 || y >= grid[0].length || grid[x][y] == -1)
            return;
        if (grid[x][y] == 2) {
            if (empty == 0) res++;
            return;
        }
        grid[x][y] = -1;
        empty--;
        dfs(grid, x + 1, y);
        dfs(grid, x - 1, y);
        dfs(grid, x, y + 1);
        dfs(grid, x, y - 1);
        grid[x][y] = 0;
        empty++;
    }
}

答案来自lee哥，没啥区别我觉得？