Unique Paths II

Unique Paths II

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

.

Note: m and n will be at most 100.

解题技巧：

该题的做法和Unique PathII的基本相同，需要进行一些特殊处理：

1.当obstacleGrid[i][j] = 1时，表示无法到达这个点，故A[i][j] = 0

2.当obstacleGrid[i -1][j] = 1时，表示无法从点(i - 1,  j)到达点(i
,  j)，故A[i][j] = A[i][j-1]

   当obstacleGrid[i][j
-1] = 1时，表示无法从点(i,  j - 1)到达点(i ,  j)，故A[i][j] = A[i-1][j]

   因此状态转移方程可写成：A[i][j]
= (1-obstacleGrid[i - 1][j] ) * A[i -1][j] + (1 - obstacleGrid[i][j - 1])
* A[i][j - 1]

代码：

int uniquePathsWithObstacles(vector< vector<int> >& obstacleGrid)
{
int m, n, flagx = 0, flagy = 0;
m = obstacleGrid.size();
n = obstacleGrid[0].size();
int A[m]
;
for(int i = 0; i < m; i ++)
{
for(int j = 0; j < n; j ++)
{
if(obstacleGrid[i][j])
{
A[i][j] = 0;
if (i == 0 && j == 0) return 0;
}
else if(i == 0 && j == 0)
{
A[i][j] = 1;
}
else if(i == 0)
{
A[i][j] = A[i][j-1];
}
else if(j == 0)
{
A[i][j] = A[i-1][j];
}
else A[i][j] = (1-obstacleGrid[i-1][j]) * A[i-1][j] + (1 - obstacleGrid[i][j-1]) * A[i][j-1];
}
}
return A[m-1][n-1];
}