leetcode_63题——Unique Paths II（动态规划）

Unique Paths II

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Question  Solution 

 

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.

 

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  Array Dynamic Programming

 

       这道题只是在前面62题的基础上加了路障，也很简单，就是在路障这个地方就把它的路径的个数直接设为0

就好了，其它的和62题没啥区别

#include<iostream>

#include <vector>

using namespace std;



int uniquePathsWithObstacles(vector<vector<int> >& obstacleGrid) {

	if(obstacleGrid.empty())

		return 0;

	int m=obstacleGrid.size();

	int n=obstacleGrid[0].size();

	int **ary1;

	ary1=new int *[m];

	for(int i=0;i<m;i++)

		ary1[i]=new int[n];

	for(int i=0;i<m;i++)

		for(int j=0;j<n;j++)

			ary1[i][j]=obstacleGrid[i][j]-1;

	for(int i=m-1;i>=0;i--)

		for(int j=n-1;j>=0;j--)

			if(ary1[i][j]!=0)

			{

				if(i==m-1&&j==n-1)

					ary1[i][j]=1;

				else if(i==m-1&&j<n-1)

					ary1[i][j]=ary1[i][j+1];

				else if(i<m-1&&j==n-1)

					ary1[i][j]=ary1[i+1][j];

				else

					ary1[i][j]=ary1[i+1][j]+ary1[i][j+1];

			}

			int last_result=ary1[0][0];



			for(int i=0;i<m;i++)

				delete[]ary1[i];

			delete[]ary1;



			return last_result;

}

int main()

{



}