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Dynamic Programming - 63. Unique Paths II

Dynamic Programming - 63. Unique Paths II

63. Unique Paths II

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

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).

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

null

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

Note: m and n will be at most 100.

Example 1:

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
**思路:**
>题目意思是棋盘左上角往右下角走,只能向下或者向右走,格子数值如果为1就不能走。题目子问题和62题一样,多的是边界情况的处理,这里多了一个如果数值为1,这个格子就不能走,换句话说,如果一个格子的上面为一个格子为1,左边的格子不为1,那么这个格子能走的种数就和左边的格子一样,就是只能从左边走过来。依次类推。而如果一个格子为1,那么能走来这个格子的种数就为0,

代码:

go:

func uniquePathsWithObstacles(obstacleGrid [][]int) int {

if obstacleGrid == nil || len(obstacleGrid) == 0 || len(obstacleGrid[0]) == 0 {
    return 0
}

m := len(obstacleGrid)
n := len(obstacleGrid[0])

if obstacleGrid[m-1][n-1] == 1 {
     return 0
}    

dp := make([][]int, m)
for i := 0; i < m; i++ {
    dp[i] = make([]int, n)
}

for i := 0; i < m; i++ {
    for j := 0; j < n; j++ {
        if obstacleGrid[i][j] == 1 {
            dp[i][j] = 0
        } else {
            if i == 0 && j == 0 {
                dp[i][j] = 1
            } else if i == 0 && j != 0 {
                dp[i][j] = dp[i][j-1]
            } else if i != 0 && j == 0 {
                dp[i][j] = dp[i-1][j]
            } else {
                dp[i][j] = dp[i][j-1] + dp[i-1][j]
            }
        }
    }
}

return dp[m-1][n-1]

}


标题:Dynamic Programming - 63. Unique Paths II
作者:michael
地址:https://www.bitbo-liuyang.com/articles/2019/07/31/1564587037307.html

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