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.
在Unique Paths的基础上稍作修改即可,在grid上为1的点,其f值为0,表示无论如何都没法到达。
class Solution { public: int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) { // Start typing your C/C++ solution below // DO NOT write int main() function if (obstacleGrid.empty() || obstacleGrid[0].empty() || obstacleGrid[0][0] == 1) { return 0; } int m = obstacleGrid.size(), n = obstacleGrid[0].size(); int f[m][n]; memset(f, 0, sizeof(int) * m * n); f[0][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++) { if (obstacleGrid[i][j] == 1) { f[i][j] = 0; } else { f[i][j] += obstacleGrid[i - 1][j] == 1 ? 0 : f[i - 1][j]; f[i][j] += obstacleGrid[i][j - 1] == 1 ? 0 : f[i][j - 1]; } } } return f[m - 1][n - 1]; } };