Unique Path II

最新推荐文章于 2025-09-07 13:58:20 发布

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最新推荐文章于 2025-09-07 13:58:20 发布

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文章标签： runtime

原文链接：http://www.cnblogs.com/amazingzoe/p/4677438.html

本文探讨了在网格中加入障碍物后，唯一路径数量的变化。通过构造2×2数组来存储路径信息，首先初始化第一列和第一行，然后计算剩余部分。此方法在3×3网格中存在一个障碍物的情况下，得出唯一路径总数为2。

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.

Analyse: Construct another 2*2 array to store path information. First initialize the first column and first row, then compute remaining parts.

Runtime: 4ms.

 1 class Solution {
 2 public:
 3     int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
 4         if(obstacleGrid[0][0] == 1) return 0;
 5         
 6         int m = obstacleGrid.size();
 7         int n = obstacleGrid[0].size();
 8         
 9         vector<vector<int> > path(m, vector<int>(n, 0));
10         path[0][0] = 1;
11         for(int i = 1; i < m; i++){ //indicate whether elements in the first column are reachable
12             if(obstacleGrid[i][0] == 0 && path[i - 1][0] == 1) path[i][0] = 1;
13             else path[i][0] = 0;
14         }
15         for(int j = 1; j < n; j++){ //indicate whether the elements in the first row are reachable
16             if(obstacleGrid[0][j] == 0 && path[0][j - 1] == 1) path[0][j] = 1;
17             else path[0][j] = 0;
18         }
19         for(int i = 1; i < m; i++){
20             for(int j = 1; j < n; j++){
21                 if(obstacleGrid[i][j] == 1) path[i][j] = 0;
22                 else path[i][j] = path[i - 1][j] + path[i][j - 1];
23             }
24         }
25         return path[m - 1][n - 1];
26     }
27 };