[Swift Weekly Contest 120]LeetCode980. 不同路径 III | Unique

发布时间：2020-12-14 05:05:39 所属栏目：百科来源：网络整理

导读：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

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

Example 2:

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

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

在二维网格?grid?上，有 4 种类型的方格：

1?表示起始方格。且只有一个起始方格。
2?表示结束方格，且只有一个结束方格。
0?表示我们可以走过的空方格。
-1?表示我们无法跨越的障碍。

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,-1]]
输出：2
解释：我们有以下两条路径：
1. (0,2)
2. (0,2)

示例 2：

输入：[[1,2]]
输出：4
解释：我们有以下四条路径： 
1. (0,3)
2. (0,3)
3. (0,3)
4. (0,3)

示例 3：

输入：[[0,0]]
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：

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

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,count+1)
44         dfs(grid,i-1,count+1)
45         dfs(grid,i,j-1,count+1)
46         dfs(grid,j+1,count+1)
47         visited[i][j] = 0
48     }
49 }

（编辑：李大同）

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