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?

Original algorithm (recursive approach) can be referred to [leet code] Pascal's Triangle.

Idea of using only O(k) extra space is reusing the same array list to form the current row of numbers.  Iterations will be from 0th row ([1]) to kth row.  And for each iteration, we need to form the current base on the previous row.

For example, in 3th row, we have [1,3,3,1].  So in the iteration of 4th row, we need only add 1 at the beginning of this array list, overwritten the 2nd element to 1+3=4, 3rd element to 3+3=6, 4th element to 3+1=4, and keep the last element remain unchanged.  Then we would have all the numbers of 4th row, i.e.[1,4,6,4,1].

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

public class Solution {
    public ArrayList<Integer> getRow(int rowIndex) {
        ArrayList<Integer> list = new ArrayList<Integer>();
        list.add(1);
        int prev, current;
        for (int i = 0; i <= rowIndex; i++) {
            prev = 1;
            for (int j = 0; j <= i; j++) {
                if (j == 0) continue;
                if (j == i) list.add(1);
                else {
                    current = list.get(j);
                    list.set(j, current + prev);
                    prev = current;
                }
            }
        }
        return list;
    }
}

public ArrayList<Integer> getRow(int rowIndex) {
    ArrayList<Integer> res = new ArrayList<Integer>();
    if(rowIndex<0)
        return res;
    res.add(1);
    for(int i=1;i<=rowIndex;i++)
    {
        for(int j=res.size()-2;j>=0;j--)
        {
            res.set(j+1,res.get(j)+res.get(j+1));
        }
        res.add(1);
    }
    return res;
}