leetcode118- Pascal's Triangle I & II- easy

I:

Given numRows, generate the first numRows of Pascal's triangle.

For example, given numRows = 5,
Return

[
     [1],
    [1,1],
   [1,2,1],
  [1,3,3,1],
 [1,4,6,4,1]
]

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?

算法，都可以根据上一层算出下一层。如果要II里面的O(k) space，可以用一个滚动的高度为2的二维数组来节省空间，因为每次只和前一次的内容有关，不需要存储太久远的内容。

实现：

题目 I

class Solution {
    public List<List<Integer>> generate(int numRows) {
        List<List<Integer>> result = new ArrayList<>();
        if (numRows < 1) {
            return result;
        }
        
        List<Integer> crt = new ArrayList<>();
        crt.add(1);
        result.add(crt);
        for (int i = 1; i < numRows; i++) {
            List<Integer> prev = result.get(result.size() - 1);
            crt = new ArrayList<Integer>();
            crt.add(1);
            for (int j = 0; j < prev.size() - 1; j++) {
                crt.add(prev.get(j) + prev.get(j + 1));
            }
            crt.add(1);
            result.add(crt);
        }
        return result;
    }
}

题目II

class Solution {
    public List<Integer> getRow(int rowIndex) {
        List<Integer> result = new ArrayList<Integer>();
        int[][] a = new int[2][rowIndex + 1];
        a[0][0] = 1;
        for (int i = 1; i <= rowIndex; i++) {
            a[i % 2][0] = 1;
            for (int j = 1; j <= i; j++) {
                a[i % 2][j] = a[(i + 1) % 2][j - 1] + a[(i + 1) % 2][j];
            }
        }
        for (int i = 0; i < rowIndex + 1; i++) {
            result.add(a[rowIndex % 2][i]);
        }
        return result;
    }
}