63. Unique Paths II

Medium

You are given an m x n integer array grid. There is a robot initially located at the top-left corner (i.e., grid[0][0]). The robot tries to move to the bottom-right corner (i.e., grid[m - 1][n - 1]). The robot can only move either down or right at any point in time.

An obstacle and space are marked as 1 or 0 respectively in grid. A path that the robot takes cannot include any square that is an obstacle.

Return the number of possible unique paths that the robot can take to reach the bottom-right corner.

The testcases are generated so that the answer will be less than or equal to 2 * 109.

Example 1:

Input: obstacleGrid = [[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

Example 2:

Input: obstacleGrid = [[0,1],[0,0]]
Output:
 1

Constraints:

• m == obstacleGrid.length

• n == obstacleGrid[i].length

• 1 <= m, n <= 100

• obstacleGrid[i][j] is 0 or 1.

解題

func uniquePathsWithObstacles(obstacleGrid [][]int) int {
    dp := obstacleGrid
    if obstacleGrid[0][0] == 0 {
        dp[0][0] = 1
    } else {
        dp[0][0] = 0
    }

    for i := 1; i < len(dp); i++ {
        if obstacleGrid[i][0] == 0 {
            dp[i][0] = dp[i - 1][0]
        } else {
            dp[i][0] = 0
        }
    }
    for i := 1; i < len(dp[0]); i++ { 
        if obstacleGrid[0][i] == 0 {
            dp[0][i] = dp[0][i - 1]
        } else {
            dp[0][i] = 0
        }
    }

    for i := 1; i < len(dp); i++ {
        for j := 1; j < len(dp[0]); j++ {
            if obstacleGrid[i][j] == 0 {
                dp[i][j] = dp[i - 1][j] + dp[i][j - 1]
            } else {
                dp[i][j] = 0
            }
        }
    }

    return dp[len(dp) - 1][len(dp[0]) - 1]
}