leetcode--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?

public class Solution {
    public List<Integer> getRow(int rowIndex) {
        ArrayList<ArrayList<Integer>> rst = new ArrayList<ArrayList<Integer>>();  
        ArrayList<Integer> first = new ArrayList<Integer>(); 
        if(rowIndex<0)
        return first;
        first.add(0, 1);  
        rst.add(first);  
  
        for (int i = 1; i < rowIndex+1; i++) {  
            ArrayList<Integer> tmp = new ArrayList<Integer>(i + 1);  
            for (int j = 0; j < i + 1; j++){  
                tmp.add(-1);  
            }  
            ArrayList<Integer> prev = rst.get(i - 1);  
            tmp.set(0, prev.get(0));  
            tmp.set(i, prev.get(i - 1));  
            for (int j = 1; j < i; j++){  
                tmp.set(j, prev.get(j - 1)+prev.get(j));  
            }  
            rst.add(tmp);  
        }  
        return rst.get(rowIndex);  
    }
}