leetcode 63. Unique Paths II
leetcode 63. Unique Paths II
A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below).
The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in the diagram below).
How many possible unique paths are there?
Above is a 7 x 3 grid. How many possible unique paths are there?An obstacle and empty space is marked as 1 and 0 respectively in the grid.
Example 1:
Input:
[
[0,0,0],
[0,1,0],
[0,0,0]
]
Output: 2
Explanation:
There is one obstacle in the middle of the 3x3 grid above.
There are two ways to reach the bottom-right corner:
- Right -> Right -> Down -> Down
- Down -> Down -> Right -> Right
Constraints:
1 <= m, n <= 100
It's guaranteed that the answer will be less than or equal to 2 * 10 ^ 9.
Solution
在每一个位置[i][j],robot要么从上面一个位置下来，要么从左边一个位置过来
所以其动态规划的状态转移方程为
dp[i][j]=dp[i-1][j]+dp[i][j-1]
但是如果[i][j]处有障碍，则该点不可达，为0
class Solution {
public:
int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
int m = obstacleGrid[0].size();
int n = obstacleGrid.size();
double dp[101][101];
for (int i = 0; i <= m; i++)
dp[0][i] = 0;
for (int i = 0; i <= n; i++)
dp[i][0] = 0;
dp[0][1] = 1;
for (int i = 1; i <= n; i++)
for (int j = 1; j <= m; j++) {
if (obstacleGrid[i - 1][j - 1] == 0)
dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
else
dp[i][j] = 0;
}
return dp[n][m];
}
};
