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

概述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 r

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].count10             {11                 if grID[i][j] == 012                 {13                     zero += 114                 }15                 if grID[i][j] == 116                 {17                     start1 = i18                     start2 = j19                 }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 ans25     }26     27     func dfs(_ grID: [[Int]],_ i:Int,_ j:Int,_ visited: [[Int]],_ count:Int)28     {29         var  visited = visited30         if i < 0 || i >= grID.count || j < 0 || j >= grID[0].count || visited[i][j] == 1 || grID[i][j] == -131         {32             return33         }34         if grID[i][j] == 235         {36             if count == zero + 137             {38                 ans += 139             }40             return41         }42         visited[i][j] = 143         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] = 048     }49 }

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