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.
思路:跟Unique Paths 类似,依然动态规划解决,不同点在于障碍处处理的细节。
代码:
int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) {
int m=obstacleGrid.size();
if(m<=0)
{
return 0;
}
int n=obstacleGrid[0].size();
if(n<=0)
{
return 0;
}
int *map=new int[n];
for(int i=n-1; i>=0; --i)
{
if(obstacleGrid[m-1][i]==1)
{
map[i]=0;
}
else
{
if(i<n-1 && map[i+1]==0)
{
map[i]=0;
}
else
{
map[i]=1;
}
}
}
for(int i=m-2; i>=0; --i)
{
for(int j=n-1; j>=0; --j)
{
if(obstacleGrid[i][j]==1)
{
map[j]=0;
}
else
{
if(j<n-1)
{
map[j]+=map[j+1];
}
}
}
}
return map[0];
}
本文介绍了一种在网格中寻找从左上角到右下角的唯一路径数量的方法,特别考虑了存在障碍的情况下如何通过动态规划解决问题。文章提供了一个具体的示例,并给出了详细的算法实现。
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