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.
1 class Solution {
2 public:
3 int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
4 if(obstacleGrid.empty() || obstacleGrid[0].empty()){
5 return 0;
6 }
7 int row = obstacleGrid.size();
8 int col = obstacleGrid[0].size();
9
10 int dp[row][col];
11
12 dp[0][0] = (obstacleGrid[0][0] == 0 ? 1 : 0);
13
14 for(int i = 1; i < row; i++){
15 dp[i][0] = ((dp[i-1][0] == 1 && obstacleGrid[i][0] == 0)? 1 : 0);
16 }
17
18 for(int j = 1; j < col; j++){
19 dp[0][j] = ((dp[0][j-1] == 1 && obstacleGrid[0][j] == 0)? 1: 0);
20 }
21
22 for(int i = 1 ; i < row; i++){
23 for(int j = 1 ; j < col;j++){
24 if(obstacleGrid[i][j] == 1){
25 dp[i][j] = 0;
26 }else{
27 dp[i][j] = dp[i-1][j]+dp[i][j-1];
28 }
29 }
30 }
31
32 return dp[row-1][col-1];
33 }
34 };