Pascal's Triangle II:节省空间杨辉三角

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 List<Integer> getRow(int rowIndex) {
	        if(rowIndex < 0) return new ArrayList<Integer>();
	        ArrayList<Integer> res = new ArrayList<Integer>();
	            res.add(1);
	        if(rowIndex == 0){            
	            return res;
	        }
	        ArrayList<Integer> temp = new ArrayList<Integer>();
	        for(int i = 0;i < rowIndex;i++){
	            temp.clear();
	            temp.add(1);
	            for(int j = 0;j< res.size()-1;j++){
	                temp.add(res.get(j)+res.get(j+1));
	            }
	            temp.add(1);
	            res.clear();
	            res.addAll(temp);
	        }
	        return res;
	    }
}

 public List<Integer> getRow(int rowIndex) {
	List<Integer> list = new ArrayList<Integer>();
	if (rowIndex < 0)
		return list;

	for (int i = 0; i < rowIndex + 1; i++) {
		list.add(0, 1);
		for (int j = 1; j < list.size() - 1; j++) {
			list.set(j, list.get(j) + list.get(j + 1));
		}
	}
	return list;
}