118. Pascal’s Triangle

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]
]

关键记录上一层的结点即pre，每一层的第i个位置，等于上一层第i-1与第i个位置之和。

	public List<List<Integer>> generate(int numRows) {
		List<List<Integer>> result = new ArrayList<>();
		if (numRows <= 0)
			return result;
		// 指向上一个三角形
		List<Integer> pre = new ArrayList<>();

		pre.add(1);
		result.add(pre);
		// i代表层数，从1开始
		for (int i = 2; i <= numRows; i++) {
			List<Integer> cur = new ArrayList<Integer>();
			// first
			cur.add(1);
			for (int j = 0; j < pre.size() - 1; j++) {
				// middle
				cur.add(pre.get(j) + pre.get(j + 1));
			}
			// last
			cur.add(1);

			result.add(cur);
			pre = cur;
		}

		return result;
	}

119. Pascal’s Triangle II

Given an index k, return the k^th 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?

获取第k行的杨辉三角，k从0开始

	public List<Integer> getRow(int rowIndex) {
		List<Integer> res = new ArrayList<Integer>();
		if (rowIndex < 0)
			return res;
		res.add(1);
		for (int i = 1; i <= rowIndex; i++) {
			// 从后往前覆盖
			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;
	}