63.Unique Paths II
===

###### tags: `Medium`,`Array`,`DP`

[63. Unique Paths II](https://leetcode.com/problems/unique-paths-ii/)

### 題目描述

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 * 10^9^.

### 範例

**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`.

### 解答
#### Python

class Solution:
    def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:
        m, n = len(obstacleGrid), len(obstacleGrid[0])
        dp = [[0] * (n + 1) for _ in range(m + 1)]
        dp[0][1] = 1
        for i in range(m):
            for j in range(n):
                dp[i + 1][j + 1] = dp[i + 1][j] + dp[i][j + 1] if obstacleGrid[i][j] == 0 else 0 
        return dp[-1][-1]

> [name=Yen-Chi Chen][time=Sun, Aug 13, 2023]

### Reference

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