Unique Paths II - Leetcode

最新推荐文章于 2021-02-24 07:42:14 发布

原创最新推荐文章于 2021-02-24 07:42:14 发布 · 391 阅读

0 ·

CC 4.0 BY-SA版权

Leetcode 专栏收录该内容

103 篇文章

订阅专栏

本文探讨了在一个含有障碍物的网格中寻找从左上角到右下角的不同路径数量的问题。介绍了一个动态规划的方法来解决这一挑战，通过一个一维数组记录到达每个格子的路径数目，并考虑了障碍物的影响。

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;
		int[] f = new int[n];

        f[0] = obstacleGrid[0][0] == 1?0:1;
		for (int i = 0; i < m; i++)
			for (int j = 0; j < n; j++) {
				if (obstacleGrid[i][j] == 1) {
					f[j] = 0;
				} else {
					if (j == 0)
						f[j] = f[j];
					else
						f[j] = f[j] + f[j - 1];
				}
			}
		return f[n - 1];
    }
}