# 63. Unique Paths II

## problem 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?

![](https://assets.leetcode.com/uploads/2018/10/22/robot_maze.png)

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
```

## algorithm thought

这题和上一题唯一的区别就是，这里在机器人的行动路上放了一些障碍。其实和上题做法没什么区别，当当前走到障碍的时候，直接把这个位置的结果赋值0。这样的好处是，既可以代表这里没有一种方案可以走到，也可以在他右边和下面得到总线路时，加上的是0。

## code

```cpp
class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
        for(int i=0;i<obstacleGrid[0].size();++i){
            if(obstacleGrid[0][i]==1)
                for(;i<obstacleGrid[0].size();++i)
                    obstacleGrid[0][i]=0;
            else
                obstacleGrid[0][i]=1;
        }
        //obstacleGrid[0][0]=0;
        for(int i=1;i<obstacleGrid.size();++i){
            if(obstacleGrid[i][0]==1||obstacleGrid[0][0]==0)
                for(;i<obstacleGrid.size();++i)
                    obstacleGrid[i][0]=0;
            else
                obstacleGrid[i][0]=1;
        }
        for(int i=1;i<obstacleGrid.size();++i){
            for(int j=1;j<obstacleGrid[0].size();++j){
                if(obstacleGrid[i][j]==1)
                    obstacleGrid[i][j]=0;
                else
                    obstacleGrid[i][j]=obstacleGrid[i-1][j]+obstacleGrid[i][j-1];
            }
        }
        return obstacleGrid.back().back();
    }
};
```

## algorithm analysis

相比于上题加入了一些处理，但是时间复杂度还是O(n²)