119 Pascal Triangle ii

Question:

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?

Code:

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

updated：

public class Solution { public List getRow(int rowIndex) { List last = new ArrayList();

    // since题目从0开始，让循环从0开始，此时，每一行的长度为i+1，最后一个值的index为i。
    for (int i = 0; i <= rowIndex; i++) {
        List<Integer> list = new ArrayList<Integer>();
        // 找最后一个位置上的index应该先找长度，然后找长度和上一重循环的对应关系。
        for (int j = 0; j <= i; j++) {
            if (j == 0 || j == i) {
                list.add(1);
            } else {
                list.add(last.get(j - 1) + last.get(j));
            }
        }
        last = list;
    }
    return last;
}

}