Description:
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.
分析：这道题目跟之前的UniquePath基本都是一样的，只是在格子中加了障碍物。基本思路是深搜+备忘录
但是不知道为什么会超时。。基本就跟答案一样，以后分析
1 int pathrec[102][102]={0}; 2 class Solution { 3 public: 4 5 int findpath(vector<vector<int> >& grid, int m, int n,int x,int y) 6 { 7 if(grid[m][n]==1) 8 return 0; 9 10 if(pathrec[m][n]!=0) 11 return pathrec[m][n]; 12 int pathnum = 0; 13 if(m==x-1 && n==y-1) 14 { 15 pathrec[m][n] = 1; 16 return 1; 17 } 18 19 if(m+1<x) pathnum+= findpath(grid,m+1,n,x,y); 20 if(n+1<y) pathnum+= findpath(grid,m,n+1,x,y); 21 pathrec[m][n] = pathnum; 22 return pathnum; 23 } 24 int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) { 25 if(obstacleGrid.empty() || obstacleGrid[0].empty()) return 0; 26 int x = obstacleGrid.size(); int y= obstacleGrid[0].size(); 27 memset(pathrec,0,sizeof(pathrec)); 28 int pathnum = findpath(obstacleGrid,0,0,x,y); 29 return pathnum; 30 } 31 };