Unique Paths II

最新推荐文章于 2024-07-25 09:04:50 发布

Nereus_Li

最新推荐文章于 2024-07-25 09:04:50 发布

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分类专栏： LeetCode

本文链接：https://blog.youkuaiyun.com/li_chihang/article/details/44338109

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本文探讨了在网格中加入障碍物后，唯一路径的数量如何计算。通过使用动态规划方法，我们解决了在存在障碍物的情况下寻找唯一路径的问题，并提供了一个具体的例子来说明解决方案。

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.

int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) {
	vector<vector<int> >Paths(obstacleGrid);
	int FG = false;
	for (int i = 0; i != Paths.size(); ++i)
	{
		if (FG)
			Paths[i][0] = 0;
		else if (obstacleGrid[i][0] == 1)
		{
			FG = true;
			Paths[i][0] = 0;
		}	
		else
			Paths[i][0] = 1;
	}
	FG = false;
	for (int j = 1; j != Paths[0].size(); ++j)
	{
		if (FG)
			Paths[0][j] = 0;
		else if (obstacleGrid[0][j] == 1)
		{
			FG = true;
			Paths[0][j] = 0;
		}
		else
			Paths[0][j] = 1;
	}
	for (int i = 1; i != Paths.size(); ++i)
	{
		for (int j = 1; j != Paths[0].size(); ++j)
		{
			if (obstacleGrid[i][j]==1)
				Paths[i][j] = 0;
			else
				Paths[i][j] = Paths[i - 1][j] + Paths[i][j - 1];
		}
	}
	return Paths[Paths.size() - 1][Paths[0].size() - 1];
}