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.

和第一题类似，dp方程还是f[i][j] = f[i-1][j] + f[i][j-1]; 只不过因为有障碍物的原因，碰到障碍物需要重置为0。

所以修改dp方程为f[i][j] = obstacleGrid[i][j] == 1 ? 0 : f[i-1][j] + f[i][j-1];

int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid){    int m=obstacleGrid.size();    int n=obstacleGrid[0].size();    int f[m][n];    f[0][0] = obstacleGrid[0][0] == 1 ? 0 : 1;    for(int i = 1; i < m; i++)        f[i][0] = obstacleGrid[i][0] == 1 ? 0 : f[i-1][0];    for(int i = 1; i < n; i++)        f[0][i] = obstacleGrid[0][i] == 1 ? 0 : f[0][i-1];    for(int i = 1; i < m; i++)        for(int j = 1; j < n; j++)            f[i][j] = obstacleGrid[i][j] == 1 ? 0 : f[i-1][j] + f[i][j-1];    return f[m-1][n-1];}