【LeetCode】63. Unique Paths II

题目：
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.
分析：
Unique Path的拓展。使用相同的状态转移方程dp[i][j] = dp[i-1][j]+dp[i][j-1]。不过要对矩阵中的每个元素进行判断，若矩阵中的元素obstacleGrid[i][j]为1,则dp[i][j]为0.
代码：

class Solution {public:    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {        int dp[101][101];        if(obstacleGrid.size() == 0)        return 0;        int m = obstacleGrid.size();        int n = obstacleGrid[0].size();        if(m == 1&&n == 1)        {            if(obstacleGrid[0][0] == 0)             return 1;             else return 0;        }        if(obstacleGrid[0][0] == 0)        dp[0][0] = 1;        else dp[0][0] = 0;        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&&j-1>=0)                dp[i][j] = dp[i-1][j]+dp[i][j-1];                else if(i-1>=0)                {                    dp[i][j] = dp[i-1][j];                }                else if(j-1>=0)                {                    dp[i][j] = dp[i][j-1];                }            }        }        return dp[m-1][n-1];    }};