Leetcode Unique Path 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 and0 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.

跟之前的Unique Path的解法类似，动态规划大法。存在一点点区别，当obstacleGrid[i][j]为1时，dp[i][j] = 0；当obstacleGrid[i][j]为0时，其状态转移方程与前一题一模一样。

代码如下：

class Solution {public:    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {        if(obstacleGrid.empty())            return 0;                int m = obstacleGrid.size();        int n = obstacleGrid[0].size();                int dp[m][n];        memset(dp,0,sizeof(int)*m*n);        dp[0][0] = 1;                    for(int i=0;i<m;i++)        {            for(int j=0;j<n;j++)            {                if(obstacleGrid[i][j] == 1)                {                    dp[i][j] = 0;                }                else                {                    if(i-1 >= 0)                        dp[i][j] += dp[i-1][j];                    if(j-1 >= 0)                        dp[i][j] += dp[i][j-1];                }            }        }        return dp[m-1][n-1];    }};