63. Unique Paths II

上一个题目的完整解题思路：
http://blog.csdn.net/daigualu/article/details/76018298

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.

注意

几个边界情况：
1.[0]
2.[0,1]
3.[0,0],[0,1]

代码实现

         //state transition equation:        //dp[i,j] = dp[i-1,j]+dp[i,j-1]         //boundary: dp[0,j]=1; dp[i,0]=1        //constraint condition:        //if(obstacleGrid[i-1,j]==1) dp[i-1,j]=0;        //if(obstacleGrid[i,j-1]==1) dp[i,j-1]=0;public class Solution {    public int UniquePathsWithObstacles(int[,] obstacleGrid) {           int m = obstacleGrid.GetUpperBound(0) + 1;            int n = obstacleGrid.GetUpperBound(1) + 1;            int[,] dp = new int[m, n];            dp[0, 0] = obstacleGrid[0, 0] == 0 ? 1 : 0;            if (obstacleGrid[m - 1, n - 1] == 1) return 0;            //two boundaries:            for (int i = 1; i < m; i++)            {                if (obstacleGrid[i, 0] == 0)                    dp[i, 0] = dp[i - 1, 0];                else dp[i, 0] = 0;            }            for (int i = 1; i < n; i++)            {                if (obstacleGrid[0, i] == 0)                    dp[0, i] = dp[0, i - 1];                else dp[0, i] = 0;            }            //dp[i][j] = dp[i-1][j]+dp[i][j-1]            for (int i = 1; i < m; i++)            {                for (int j = 1; j < n; j++)                {                    if (obstacleGrid[i - 1, j] == 1)                        dp[i - 1, j] = 0;                    if (obstacleGrid[i, j - 1] == 1)                        dp[i, j - 1] = 0;                    dp[i, j] = dp[i - 1, j] + dp[i, j - 1];                }            }            return dp[m - 1, n - 1];    }}