[LeetCode] Unique Paths II 解题报告-优快云博客

本文探讨了在网格中加入障碍物后，求解唯一路径的数量。通过转移方程和动态规划的方法，解决了该问题，并给出了具体实现代码。

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 3×3 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.

» Solve this problem

[解题思路]

和Unique Path一样的转移方程：

Step[i][j] = Step[i-1][j] + Step[i][j-1] if Array[i][j] ==0
or = 0 if Array[i][j] =1

[Code]

1:       int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) {  
2:            int m = obstacleGrid.size();  
3:            if(m ==0) return 0;  
4:            int n = obstacleGrid[0].size();  
5:            if(obstacleGrid[0][0] ==1) return 0;  
6:            vector<int> maxV(n,0);  
7:            maxV[0] =1;  
8:            for(int i =0; i<m; i++)  
9:            {  
10:                 for(int j =0; j<n; j++)  
11:                 {  
12:                      if(obstacleGrid[i][j] ==1)  
13:                           maxV[j]=0;  
14:                      else if(j >0)  
15:                           maxV[j] = maxV[j-1]+maxV[j];  
16:                 }  
17:            }  
18:            return maxV[n-1];  
19:       }