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:

1. Right -> Right -> Down -> Down
2. 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];
	}
};

参考链接

leetcode