## Description

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?

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

## Solution

### 动态规划（Dynamic Programming）

$$f(x, y) = f(x-1, y) + f(x, y-1)$$

### Example

0, 0, 0, 0       1, 1, 1, 1
0, 0, 0, 0  ==>  1, 2, 3, 4
0, 0, 0, 0       1, 3, 6, 10

0, 0, 0, 0       1, 1, 1, 1
0, 0, 1, 0  ==>  1, 2, 0, 1
0, 0, 0, 0       1, 3, 3, 4

### Code Here

 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 func uniquePathsWithObstacles(obstacleGrid [][]int) int { row := len(obstacleGrid) if row == 0 { return 0 } col := len(obstacleGrid[0]) if col == 0 || obstacleGrid[0][0] == 1 { return 0 } var left, up int var ways int // 标记第一格到达路径数量为1 obstacleGrid[0][0] = 1 j := 1 // 从第二个格子开始 for i := 0; i < row; i++ { for ; j < col; j++ { // 如果为障碍，标记当前格子达到路径数量为 0 if obstacleGrid[i][j] == 1 { obstacleGrid[i][j] = 0 continue } // 表示当前格子到达路径数量 // 它等于左边格子路径数量+上边格子路径数量 ways = 0 left = j - 1 if left >= 0 { // 加上左边格子路径数量 // ways += leftWays ways = obstacleGrid[i][left] } up = i - 1 if up >= 0 { // 加上上边格子路径数量 ways += obstacleGrid[up][j] } // 保存当前格子路径数量 obstacleGrid[i][j] = ways } j = 0 } // 返回最后一个格子路径数量 return obstacleGrid[row-1][col-1] }