[Swift]LeetCode63. 不一样路径 II | Unique Paths II

时间 2019-11-11

标签 swift leetcode63 leetcode 不一样路径 unique paths 栏目 Swift 繁體版

原文原文链接

★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤微信公众号：山青咏芝（shanqingyongzhi）
➤博客园地址：山青咏芝（https://www.cnblogs.com/strengthen/）
➤GitHub地址：https://github.com/strengthen/LeetCode
➤原文地址：http://www.javashuo.com/article/p-qiaadmsn-me.html
➤若是连接不是山青咏芝的博客园地址，则多是爬取做者的文章。
➤原文已修改更新！强烈建议点击原文地址阅读！支持做者！支持原创！
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★html

A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below).git

The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in the diagram below).github

Now consider if some obstacles are added to the grids. How many unique paths would there be?微信

An obstacle and empty space is marked as 1 and 0 respectively in the grid.ide

Note: m and n will be at most 100.spa

Example 1:code

Input:
[
  [0,0,0],
  [0,1,0],
  [0,0,0]
]
Output: 2
Explanation:
There is one obstacle in the middle of the 3x3 grid above.
There are two ways to reach the bottom-right corner:
1. Right -> Right -> Down -> Down
2. Down -> Down -> Right -> Right

一个机器人位于一个 m x n 网格的左上角（起始点在下图中标记为“Start” ）。htm

机器人每次只能向下或者向右移动一步。机器人试图达到网格的右下角（在下图中标记为“Finish”）。blog

如今考虑网格中有障碍物。那么从左上角到右下角将会有多少条不一样的路径？get

网格中的障碍物和空位置分别用 1 和 0 来表示。

说明：m 和 n 的值均不超过 100。

示例 1:

输入:
[
  [0,0,0],
  [0,1,0],
  [0,0,0]
]
输出: 2
解释:
3x3 网格的正中间有一个障碍物。
从左上角到右下角一共有  条不一样的路径：
1. 向右 -> 向右 -> 向下 -> 向下
2. 向下 -> 向下 -> 向右 -> 向右2

12ms

 1 class Solution {
 2     func uniquePathsWithObstacles(_ obstacleGrid: [[Int]]) -> Int {
 3         let rowCount = obstacleGrid.count
 4         let colCount = obstacleGrid.first?.count ?? 0
 5         guard rowCount > 0, colCount > 0 else { return 0 }
 6         var paths = Array(repeating: Array(repeating: 0, count: colCount), count: rowCount)
 7         
 8         if obstacleGrid[0][0] != 1 {
 9             paths[0][0] = 1
10         }
11         for i in 1..<rowCount {
12             if obstacleGrid[i][0] != 1 {
13                 paths[i][0] = paths[i-1][0]
14             }
15         }
16         
17         for i in 1..<colCount {
18             if obstacleGrid[0][i] != 1 {
19                 paths[0][i] = paths[0][i-1]
20             }
21         }
22         
23         for r in 1..<rowCount {
24             for c in 1..<colCount {
25                 if obstacleGrid[r][c] != 1 {
26                     paths[r][c] = paths[r-1][c] + paths[r][c-1]
27                 } else {
28                     paths[r][c] = 0
29                 }
30             }
31         }
32         
33         return paths[rowCount-1][colCount-1]
34     }
35 }

16ms

 1 class Solution {
 2     func uniquePathsWithObstacles(_ obstacleGrid: [[Int]]) -> Int {
 3         
 4         let m = obstacleGrid.count
 5         if m == 0 { return 0 }
 6         let n = obstacleGrid[0].count
 7         if n == 0 { return 0 }
 8                 
 9         var f = [[Int]](repeating: [Int](repeating: 0, count: n), count: m)
10         
11         for i in 0..<m {
12             if obstacleGrid[i][0] == 1 {
13                 break
14             } else {
15                 f[i][0] = 1
16             }
17         }
18         
19         for i in 0..<n {
20             if obstacleGrid[0][i] == 1 {
21                 break
22             } else {
23                 f[0][i] = 1
24             }
25         }
26         
27         for i in 1..<m {
28             for j in 1..<n {
29                 
30                 if obstacleGrid[i][j] == 1 {
31                     f[i][j] = 0
32                 } else {
33                     f[i][j] = f[i - 1][j] + f[i][j - 1]
34                 }
35             }
36         }
37         
38         return f[m - 1][n - 1]
39     }
40 }

16ms

 1 class Solution {
 2     func uniquePathsWithObstacles(_ obstacleGrid: [[Int]]) -> Int {
 3         guard obstacleGrid.count > 0 && obstacleGrid[0].count > 0 else {
 4             return 0
 5         }
 6         var m = obstacleGrid.count, n = obstacleGrid[0].count
 7         var dp = [[Int]](repeating: [Int](repeating: 0, count: n), count: m)
 8         for i in 0..<m {
 9             if obstacleGrid[i][0] == 1 {
10                 for k in i..<m {
11                     dp[k][0] = 0
12                 }
13                 break
14             } else {
15                 dp[i][0] = 1
16             }
17         }
18         for j in 0..<n {
19             if obstacleGrid[0][j] == 1 {
20                 for k in j..<n {
21                     dp[0][k] = 0
22                 }
23                 break
24             } else {
25                 dp[0][j] = 1
26             }
27             
28         }
29 
30         for i in 1..<m {
31             for j in 1..<n {
32                 if obstacleGrid[i][j] == 1 {
33                     dp[i][j] = 0
34                 } else {
35                     dp[i][j] = dp[i - 1][j] + dp[i][j - 1]
36                 }
37                 
38             }
39         }
40 
41         return dp[m - 1][n - 1]
42     }
43 }

20ms

 1 class Solution {
 2     func uniquePathsWithObstacles(_ obstacleGrid: [[Int]]) -> Int {
 3         guard obstacleGrid.count > 0 && obstacleGrid[0].count>0 else {
 4             return 0
 5         }
 6  
 7         let rowCount = obstacleGrid.count
 8         let colCount = obstacleGrid[0].count
 9         var dp = [[Int]](repeating:[Int](repeating:1, count:colCount), count:rowCount)
10         var occupiedRow = false
11         var occupiedCol = false
12         for row in 0..<rowCount {
13            if obstacleGrid[row][0] == 1 || occupiedRow {
14                occupiedRow = true
15                dp[row][0] = 0
16            } 
17         }
18             
19         for col in 0..<colCount {
20            if obstacleGrid[0][col] == 1 || occupiedCol {
21                occupiedCol = true
22                dp[0][col] = 0
23            } 
24         }
25         
26         for i in 1..<rowCount {
27             for j in 1..<colCount{
28                 if obstacleGrid[i][j] == 1 {
29                     dp[i][j] = 0
30                 } else {
31                     dp[i][j] =  (obstacleGrid[i-1][j] == 1 ? 0 : dp[i-1][j]) + (obstacleGrid[i][j-1] == 1 ? 0 : dp[i][j-1])
32                 }    
33             }
34         }
35         
36         return dp[rowCount-1][colCount-1]    
37     }
38 }

24ms

 1 class Solution {
 2     func uniquePathsWithObstacles(_ obstacleGrid: [[Int]]) -> Int {
 3         var Result = Array(repeating: Array(repeating: 0, count:obstacleGrid[0].count)
 4         , count: obstacleGrid.count)
 5         if obstacleGrid.count == 0 || obstacleGrid[0].count == 0 || obstacleGrid[0][0] == 1 {
 6             return 0
 7         }
 8         // 设置边界值，碰到障碍物后都无路径
 9         for i in 0..<obstacleGrid.count {
10              if obstacleGrid[i][0] == 1 {
11                  break
12              }
13              Result[i][0] = 1
14         }
15         // 设置边界值，碰到障碍物后都无路径
16         for j in 0..<obstacleGrid[0].count {
17             if obstacleGrid[0][j] == 1 {
18                 break
19             }
20             Result[0][j] = 1
21         }
22 
23         // 动态规划求出路径（迭代法）
24         for i in 1..<obstacleGrid.count {
25             for j in 1..<obstacleGrid[0].count {
26                 if obstacleGrid[i][j] == 0 {
27                     Result[i][j] = Result[i-1][j] + Result[i][j-1]
28                 } else { // 障碍物无路劲
29                     Result[i][j] = 0
30                 }
31             }
32         }
33         return Result[Result.count - 1][Result[0].count - 1]
34     }
35 }

24ms

 1 class Solution {
 2     func uniquePathsWithObstacles(_ obstacleGrid: [[Int]]) -> Int {
 3         let R = obstacleGrid.count
 4         let C = obstacleGrid[0].count
 5         var obstacleGrid = obstacleGrid
 6 
 7         // If the starting cell has an obstacle, then simply return as there would be
 8         // no paths to the destination.
 9         if obstacleGrid[0][0] == 1 {
10             return 0
11         }
12 
13         // Number of ways of reaching the starting cell = 1.
14         obstacleGrid[0][0] = 1
15 
16         // Filling the values for the first column
17         for i in 1..<R {
18             obstacleGrid[i][0] = (obstacleGrid[i][0] == 0 && obstacleGrid[i - 1][0] == 1) ? 1 : 0
19         }
20 
21         // Filling the values for the first row
22         for i in 1..<C {
23             obstacleGrid[0][i] = (obstacleGrid[0][i] == 0 && obstacleGrid[0][i - 1] == 1) ? 1 : 0
24         }
25 
26         // Starting from cell(1,1) fill up the values
27         // No. of ways of reaching cell[i][j] = cell[i - 1][j] + cell[i][j - 1]
28         // i.e. From above and left.
29         for i in 1..<R {
30             for j in 1..<C {
31                 if obstacleGrid[i][j] == 0 {
32                     obstacleGrid[i][j] = obstacleGrid[i - 1][j] + obstacleGrid[i][j - 1]
33                 } else {
34                     obstacleGrid[i][j] = 0
35                 }
36             }
37         }
38 
39         // Return value stored in rightmost bottommost cell. That is the destination.
40         return obstacleGrid[R - 1][C - 1]
41     }
42 }