[Swift Weekly Contest 120]LeetCode980. 不同路徑 III | Unique Paths III
阿新 • • 發佈:2019-01-20
types leet can Dimension || Once col wal clas
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
在二維網格 grid
上,有 4 種類型的方格:
1
表示起始方格。且只有一個起始方格。2
表示結束方格,且只有一個結束方格。0
表示我們可以走過的空方格。-1
表示我們無法跨越的障礙。
返回在四個方向(上、下、左、右)上行走時,從起始方格到結束方格的不同路徑的數目,每一個無障礙方格都要通過一次。
示例 1:
輸入:[[1,0,0,0],[0,0,0,0],[0,0,2,-1]] 輸出:2 解釋:我們有以下兩條路徑: 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:
輸入:[[1,0,0,0],[0,0,0,0],[0,0,0,2]] 輸出:4 解釋:我們有以下四條路徑: 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:
輸入:[[0,1],[2,0]] 輸出:0 解釋: 沒有一條路能完全穿過每一個空的方格一次。 請註意,起始和結束方格可以位於網格中的任意位置。
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, 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 }
[Swift Weekly Contest 120]LeetCode980. 不同路徑 III | Unique Paths III