[LeetCode] Unique Paths II 解题报告

本文介绍了一个基于动态规划解决的问题——在包含障碍物的网格中计算从左上角到右下角的不同路径数量。通过更新一个一维数组来记录可达位置的数量,并在遇到障碍时将路径数量置零。

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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.
» 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: }




转载于:https://www.cnblogs.com/codingtmd/archive/2013/01/11/5078939.html

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