Pascal's Triangle II

Given an index k, return the kth row of the Pascal's triangle.

For example, given k = 3,
Return [1,3,3,1].

Note:

Could you optimize your algorithm to use only O(k) extra space?

 This is extension problem of Pascal's Triangle.
    In previous problem, we solve it select pair by order of front to back.
    so we need use Extra ArrayList to store
    value from previous level in order to generate value for next level
    that value of position i=pre[i]+pre[i+1].
    In general, we can solve problem by selecting pair from back to front that in order of
    right -> left. position j start from back => 
    position i+1=res[i]+res[i+1]
    there is no extra space used by this method.

public class Solution {   public ArrayList<Integer> getRow(int rowIndex) {        ArrayList<Integer> res = new ArrayList<Integer>();        //check for special case        if(rowIndex==0) return res;        res.add(1);        for(int i=1;i<=rowIndex;i++){            int size=res.size();            for(int j=size-2;j>=0;j--){                //calculate selected pair from back to front                res.set(j+1,res.get(j)+res.get(j+1));            }            res.add(1);        }        return res;    }}