LeetCode Unique Paths II

最新推荐文章于 2025-09-06 17:22:57 发布

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本文介绍了一种基于动态规划的方法来解决在二维网格中寻找从左上角到右下角的路径数量的问题，当网格中存在障碍物时。文章提供了一个Java实现示例，并详细解释了如何处理边界条件。

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.

Note: m and n will be at most 100.

题意：

此题是Unique Paths的一个变种，就是在二维数组中增加一些特殊的障碍，但是其实归根到底都是DP(动态规划)的类型。考虑几种特殊情况就行了，首先这里增加了一个obstacleGrid数组，这个数组中存储有每个位置上的值，1代表有障碍，0代表没有障碍；那么我另外又开了一个二维数组用来保存走到每一个位置的种树。但是这里要考虑的是，在第一行和第一列中，如果一旦一个位置出现了1，那么后面的所有格子都将是0，这里要特别注意。另外格子和之前考虑的一样，都是左边和上边的那两个格子的值相加。

public static int uniquePathsWithObstacles(int[][] obstacleGrid)
	{
		int m = obstacleGrid.length;
		int n = obstacleGrid[0].length;
		int[][] nums = new int[m][n];
		if(obstacleGrid[0][0] == 1)
			return 0;
		nums[0][0] = 1;
		for(int i = 0; i < m; i++)
		{
			for(int j = 0; j < n; j++)
			{
				if(i == 0 && j > 0)
				{
					if(obstacleGrid[i][j] == 0 && obstacleGrid[i][j - 1] != 1 && nums[i][j-1] != 0)
						nums[i][j] = 1;
					else if(obstacleGrid[i][j] == 0 && obstacleGrid[i][j - 1] == 1)
						nums[i][j] = 0;
					else if(obstacleGrid[i][j] == 1)
						nums[i][j] = 0;
				}
				if(i > 0 && j == 0)
				{
					if(obstacleGrid[i][j] == 0 && obstacleGrid[i-1][j] != 1 && nums[i-1][j] != 0)
						nums[i][j] = 1;
					else if(obstacleGrid[i][j] == 0 && obstacleGrid[i-1][j] == 1)
						nums[i][j] = 0;
					else if(obstacleGrid[i][j] == 1)
						nums[i][j] = 0;
				}
				if(i > 0 && j > 0)
				{
					if(obstacleGrid[i][j] == 1)
						nums[i][j] = 0;
					else if(obstacleGrid[i][j] != 1)
						nums[i][j] = nums[i-1][j] + nums[i][j-1];
				}
			}
		}
		return nums[m-1][n-1];
	}