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:
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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:
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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]
}