LeetCode Unique Paths II

最新推荐文章于 2019-05-17 12:08:12 发布

陈琼

最新推荐文章于 2019-05-17 12:08:12 发布

阅读量363

点赞数

分类专栏： LeetCode 文章标签： leetcode dp

本文链接：https://blog.youkuaiyun.com/xuyilanqi/article/details/40824849

版权

LeetCode 专栏收录该内容

21 篇文章

订阅专栏

探讨如何在网格中存在障碍的情况下计算从左上角到右下角的唯一路径数量。采用动态规划方法，考虑网格大小最多为100x100。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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.

public class Solution {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        int m=obstacleGrid.length;
        int n=obstacleGrid[0].length;
        if(obstacleGrid==null)
        	return 0;
        if(m<=0||n<=0)
			return 0;
        if(obstacleGrid[0][0]==1)
        	return 0;
        if(obstacleGrid[m-1][n-1]==1)
        	return 0;
        int[][] f=new int[m+1][n+1];
        for(int i=0;i<=m;i++)
			f[i][n]=0;
		for(int i=0;i<=n;i++)
			f[m][i]=0;
		for(int i=0;i<m;i++)
			for(int j=0;j<n;j++)
			{
				if(obstacleGrid[i][j]==1)
					f[i][j]=0;
			}
		for(int i=m-1;i>=0;i--)
			for(int j=n-1;j>=0;j--)
			{
				if(obstacleGrid[i][j]==0)
					f[i][j]=f[i+1][j]+f[i][j+1];
				else f[i][j]=0;
				f[m-1][n-1]=1;
			}
		return f[0][0];
    }
}