[LeetCode117]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.
Analysis:
there is a little difference with the Unique Paths
DP functions
dp[i]][j] = dp[i-1][j]+dp[i][j-1] if obstacle[i][j] == 0
= 0 if obstacle[i][j] == 1
java
public int uniquePathsWithObstacles(int[][] obstacleGrid) {
       if(obstacleGrid[0][0] ==1) return 0;
		int m = obstacleGrid.length;
		int n = obstacleGrid[0].length;
		int[] step = new int[n];
		step[0] = 1;
		for(int i=0;i<m;i++){
			for(int j=0;j<n;j++){
				if(obstacleGrid[i][j]==1)
					step[j] = 0;
				else if(j>0){
					step[j] = step[j-1]+step[j];
				}
			}
		}
		return step[n-1];
    }
c++
int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) {
        int m = obstacleGrid.size();
    if(m==0) return 0;
    int n = obstacleGrid[0].size();
    if(obstacleGrid[0][0] == 1) return 0;
    vector<int> maxpath(n,0);
    maxpath[0] = 1;
    for(int i=0; i< m; i++){
        for(int j=0;j<n;j++){
            if(obstacleGrid[i][j] == 1)
                maxpath[j] = 0;
            else if(j>0)
                maxpath[j] = maxpath[j]+maxpath[j-1];
        }
    }
    return maxpath[n-1];
    }

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