Unique Paths II <leetcode>

Follow up for "Unique Paths":

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.

For example,

There is one obstacle in the middle of a 3x3 grid as illustrated below.

[
  [0,0,0],
  [0,1,0],
  [0,0,0]
]

The total number of unique paths is 2.

算法：一开始理解错了，以为到一个点可以从上下左右，写了一个非常复杂的代码，后来自己感觉不对，上网上一查才发现，原来还是到一个点只能从上或左，那这道题就非常简单了，思路和第一个版本是一样的，只是增加了障碍物，所以计算的时候，考虑一下障碍物就行了，代码如下（动态规划）：

class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) {
        vector<vector<int>>  f(obstacleGrid.size(),vector<int>(obstacleGrid[0].size()));
        f[0][0]=obstacleGrid[0][0]==0?1:0;
        for(int i=1;i<f.size();i++)
        {
            f[i][0]=obstacleGrid[i][0]==0?f[i-1][0]:0;
        }
        for(int i=1;i<obstacleGrid[0].size();i++)
        {
            f[0][i]=obstacleGrid[0][i]==0?f[0][i-1]:0;
        }
        
        for(int i=1;i<obstacleGrid.size();i++)
        {
            for(int j=1;j<obstacleGrid[0].size();j++)
            {
                f[i][j]=obstacleGrid[i][j]==0?(f[i-1][j]+f[i][j-1]):0;
            }
        }
        return f[f.size()-1][f[0].size()-1];
    }
};