【LeetCode】63. Unique Paths II（C++）

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

题目：

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: m and n will be at most 100.
Example 1:

理解：

和上面的题一样，只是如果这个位置为1，则把dp赋为0即可。

实现：

还是按昨天的思路实现的。
只是这种思路空间复杂度高，为 $O(mn)$
为了代码短，多申请了一行一列

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];
	}
};

节省一下空间复杂度

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];
	}
};