地址：https://leetcode.com/problems/unique-paths-ii/

題目：

A robot is located at the top-left corner of a $m * n$ 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).

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.
Note

理解：

和上面的題一樣，只是如果這個位置為1，則把dp賦為0即可。

實現：

還是按昨天的思路實現的。
只是這種思路空間複雜度高，為 $O(mn)$

class Solution {
public:
	int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
		if (obstacleGrid[0][0]) return 0;
		int n = obstacleGrid.size(), m = obstacleGrid[0].size();
		vector<vector<int>> dp(n + 1, vector<int>(m + 1, 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])
					dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
			}
		return dp[n][m];
	}
};

節省一下空間複雜度

class Solution {
public:
	int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
		if (obstacleGrid[0][0]) return 0;
		int n = obstacleGrid.size(), m = obstacleGrid[0].size();
		vector<int> dp(m, 0);
		for (int j = 0; j < m; ++j) {
			if (!obstacleGrid[0][j])
				dp[j] = 1;
			else
				break;
		}
		for (int i = 1; i < n; ++i) {
			if (obstacleGrid[i][0])
				dp[0] = 0;
			for (int j = 1; j < m; ++j) {
				if (!obstacleGrid[i][j]) {
					dp[j] += dp[j - 1];
				}
				else
					dp[j] = 0;
			}
		}
		return dp[m-1];
	}
};