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★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★ On a 2-dimensional
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:
在二维网格
返回在四个方向(上、下、左、右)上行走时,从起始方格到结束方格的不同路径的数目,每一个无障碍方格都要通过一次。 示例 1: 输入:[[1,0,0,0],[0,0,0,0],[0,0,2,-1]] 输出:2 解释:我们有以下两条路径: 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) 示例 2: 输入:[[1,0,0,0],[0,0,0,0],[0,0,0,2]] 输出:4 解释:我们有以下四条路径: 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) 示例 3: 输入:[[0,1],[2,0]] 输出:0 解释: 没有一条路能完全穿过每一个空的方格一次。 请注意,起始和结束方格可以位于网格中的任意位置。 提示:
32ms 1 class Solution { 2 var zero:Int = 0 3 var ans:Int = 0 4 func uniquePathsIII(_ grid: [[Int]]) -> Int { 5 var start1:Int = 0 6 var start2:Int = 0 7 for i in 0..<grid.count 8 { 9 for j in 0..<grid[0].count 10 { 11 if grid[i][j] == 0 12 { 13 zero += 1 14 } 15 if grid[i][j] == 1 16 { 17 start1 = i 18 start2 = j 19 } 20 } 21 } 22 var visited:[[Int]] = [[Int]](repeating:[Int](repeating:0,count:grid[0].count),count:grid.count) 23 dfs(grid, start1, start2, visited, 0) 24 return ans 25 } 26 27 func dfs(_ grid: [[Int]],_ i:Int,_ j:Int,_ visited: [[Int]],_ count:Int) 28 { 29 var visited = visited 30 if i < 0 || i >= grid.count || j < 0 || j >= grid[0].count || visited[i][j] == 1 || grid[i][j] == -1 31 { 32 return 33 } 34 if grid[i][j] == 2 35 { 36 if count == zero + 1 37 { 38 ans += 1 39 } 40 return 41 } 42 visited[i][j] = 1 43 dfs(grid, i+1, j, visited, count+1) 44 dfs(grid, i-1, j, visited, count+1) 45 dfs(grid, i, j-1, visited, count+1) 46 dfs(grid, i, j+1, visited, count+1) 47 visited[i][j] = 0 48 } 49 }
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