Unique Paths II
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.
1 class Solution { 2 public: 3 int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) { 4 int m = obstacleGrid.size(); 5 if(m == 0) return 0; 6 int n = obstacleGrid[0].size(); 7 8 vector<int> row(n,0); 9 vector<vector<int>> record(m,row); 10 11 bool obstacle = false; 12 for(int i = n-1; i >= 0; i--){ 13 if(obstacle) 14 record[m-1][i] = 0; 15 else if(obstacleGrid[m-1][i] == 1){ 16 obstacle = true; 17 record[m-1][i] = 0; 18 }else record[m-1][i] = 1; 19 } 20 obstacle = false; 21 for(int i = m-1; i >= 0; i--){ 22 if(obstacle) 23 record[i][n-1] = 0; 24 else if(obstacleGrid[i][n-1] == 1){ 25 obstacle = true; 26 record[i][n-1] = 0; 27 }else record[i][n-1] = 1; 28 29 } 30 for(int i = m-2; i >= 0; i--) 31 for(int j = n-2; j >= 0; j--){ 32 if(obstacleGrid[i][j] == 1) record[i][j] = 0; 33 else record[i][j] = record[i+1][j] + record[i][j+1]; 34 } 35 return record[0][0]; 36 } 37 };
动态规划,时间复杂度O(m*n), 空间复杂度O(m*n)。