[Leetcode]-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.

Hide Tags Array Dynamic Programming
Hide Similar Problems (M) Unique Paths
题目：这道题目和Unique Paths类似，只不过在其中增加了障碍物，obstacleGrid[i][j]中为0表示无障碍，obstacleGrid[i][j]为1表示为障碍物不能从这里通过。

思路：
分配path[obstacleGridRowSize][obstacleGridColSize]二维数组，用来存放能到当前位置的路径数目
　　１、设置首行，只要已发现出现obstacleGrid[0][i]第一个有1，则将后面的path[0][j]全部设置为0；
　　2、设置首列，只要已发现出现obstacleGrid[i][0]第一个有1，则将后面的path[j][0]全部设置为0；
　　3、循环其他所有位置：
　　　　　１、当发现 obstacleGrid[i][j]为1，说明此路不通，对应的path[i][j]设置为0
　　　　　2、否则就计算path[i][j] = path[i-1][j] + path[i][j-1];

int uniquePathsWithObstacles(int** obstacleGrid, int obstacleGridRowSize, int obstacleGridColSize) {    int m = obstacleGridRowSize;    int n = obstacleGridColSize;    int** path =(int**)malloc(sizeof(int*) * m);    for(int i=0;i<m;i++)        path[i] = (int*)malloc(sizeof(int)*n);    for(int i=0;i<m;i++)    {        if(1 == obstacleGrid[i][0])        {            for(int j=i;j<m;j++)                path[j][0] = 0;            break;        }        else            path[i][0] = 1;    }      for(int i=0;i<n;i++)    {        if(1 == obstacleGrid[0][i])        {            for(int j=i;j<n;j++)                path[0][j] = 0;            break;        }        else            path[0][i] = 1;    }    for(int i=1;i<m;i++)    {        for(int j=1;j<n;j++)        {            if(1 == obstacleGrid[i][j])                path[i][j] = 0;            else                path[i][j] = path[i-1][j] + path[i][j-1];        }    }    int r = path[m-1][n-1];    for(int i=0;i<m;i++)        free(path[i]);    free(path);    return r;}