Unique Paths II 解答

最新推荐文章于 2024-07-30 16:17:15 发布

转载最新推荐文章于 2024-07-30 16:17:15 发布 · 50 阅读

0 ·

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/ireneyanglan/p/4822785.html

本文介绍了一种解决网格中存在障碍物时寻找从起点到终点的唯一路径数量的方法。通过动态规划，考虑了障碍物对路径的影响。

Question

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.

Solution

Similar with "Unique Paths", there are two differences to be considered:

1. for dp[0][i] and dp[i][0], if there exists previous element which equals to 1, then the rest elements are all unreachable.

2. for dp[i][j], if it equals to 1, then it's unreachable.

 1 public class Solution {
 2     public int uniquePathsWithObstacles(int[][] obstacleGrid) {
 3         if (obstacleGrid == null)
 4             return 0;
 5         int m = obstacleGrid.length, n = obstacleGrid[0].length;
 6         int[][] dp = new int[m][n];
 7         // Check first element
 8         if (obstacleGrid[0][0] == 1)
 9             return 0;
10         else
11             dp[0][0] = 1;
12         // Left column
13         for (int i = 1; i < m; i++) {
14             if (obstacleGrid[i][0] == 1)
15                 dp[i][0] = 0;
16             else
17                 dp[i][0] = dp[i - 1][0];
18         }
19         // Top row
20         for (int i = 1; i < n; i++) {
21             if (obstacleGrid[0][i] == 1)
22                 dp[0][i] = 0;
23             else
24                 dp[0][i] = dp[0][i - 1];
25         }
26         // Inside
27         for (int i = 1; i < m; i++) {
28             for (int j = 1; j < n; j++) {
29                 if (obstacleGrid[i][j] == 1)
30                     dp[i][j] = 0;
31                 else
32                     dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
33             }
34         }
35         return dp[m - 1][n - 1];
36     }
37 }

转载于:https://www.cnblogs.com/ireneyanglan/p/4822785.html