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?