# 【LeetCode】 63. Unique Paths II
## Description
> 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?
> 有一個機器人在一個m*n的網格裡的最左上角。
> 它每次只能往下或往右走一格,它的終點是最右下角。
> 現在需要考慮在網格上可能有障礙物,它有幾種不同的路徑?

> An obstacle and empty space is marked as 1 and 0 respectively in the grid.
> Note: m and n will be at most 100.
> 障礙物和空格在網格上分別用1和0表示。
> 提示:m和n最多只有100。
## Example:
```
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
* 使用DP解,`自己這格的路徑` = `上面那格的路徑` + `左邊那格的路徑`
* 再設定特殊狀況:障礙物路徑數必定為`0`、超出邊界路徑數=`0`、起始點=`0`。
### Code
```C++=1
class Solution {
public:
int getPath(vector<vector<int>>& obstacleGrid, int *table[], int x, int y)
{
//out of board
if(x<0||y<0) return 0;
//have obstacle
if(obstacleGrid[y][x]==1) return 0;
//start point
if(x==0&&y==0) return 1;
//recursion get answer
if(table[y][x] == -1) table[y][x] = getPath(obstacleGrid,table,x-1,y) + getPath(obstacleGrid,table,x,y-1);
return table[y][x];
}
int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
//get size of grid.
int s_y = obstacleGrid.size();
int s_x = obstacleGrid[0].size();
//make DP table
int **table = new int*[s_y];
for(int i=0;i<s_y;i++)
table[i] = new int[s_x];
for(int i=0;i<s_y;i++)
for(int j=0;j<s_x;j++)
table[i][j] = -1;
return getPath(obstacleGrid,table,s_x-1,s_y-1);
}
};
```
###### tags: `LeetCode` `C++`