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?

class Solution 
{
public:
    vector<int> getRow(int rowIndex) 
    {
        vector<int> pre;
        vector<int> res;
        res.push_back(1);
        for(int i=1; i<=rowIndex; i++)
        {
            pre = res;
            res.clear();
            res.push_back(1);
            for(int j=1; j<pre.size(); j++)
                res.push_back(pre[j]+pre[j-1]);
            res.push_back(1);
        }
        return res;
    }
};

LeetCode--Pascal's Triangle II