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.

思路：思路同Unique Path，只不过需要加入些许限制，注意，如果第一行和第一列的某一个值为0， 则该行以及该列后面的值都会为0

代码一：

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

// 动规，滚动数组
// 时间复杂度O(n^2)，空间复杂度O(n)
class Solution {
public:
	int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) {
		const int m = obstacleGrid.size();
		const int n = obstacleGrid[0].size();
		if (obstacleGrid[0][0] || obstacleGrid[m-1][n-1]) return 0;
		vector<int> f(n, 0);
		f[0] = obstacleGrid[0][0] ? 0 : 1;
		for (int i = 0; i < m; i++)
			for (int j = 0; j < n; j++)
				f[j] = obstacleGrid[i][j] ? 0 : (j == 0 ? 0 : f[j - 1]) + f[j];
		return f[n - 1];
	}
};