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119. Pascal's Triangle II

题目

Given an integer rowIndex, return the rowIndexth row of the Pascal's triangle.

Notice that the row index starts from 0.

In Pascal's triangle, each number is the sum of the two numbers directly above it.

Follow up:

Could you optimize your algorithm to use only O(k) extra space?

Example 1:

Input: rowIndex = 3
Output: [1,3,3,1]

Example 2:

Input: rowIndex = 0
Output: [1]

Example 3:

Input: rowIndex = 1
Output: [1,1]

Constraints:

0 <= rowIndex <= 33

题目大意

给定一个非负索引 k，其中 k ≤ 33，返回杨辉三角的第 k 行。

解题思路

题目中的三角是杨辉三角，每个数字是 (a+b)^n 二项式展开的系数。题目要求我们只能使用 O(k) 的空间。那么需要找到两两项直接的递推关系。由组合知识得知：

$$\begin{aligned}C_{n}^{m} &= \frac{n!}{m!(n-m)!} \C_{n}^{m-1} &= \frac{n!}{(m-1)!(n-m+1)!}\end{aligned}$$

于是得到递推公式：

$$C_{n}^{m} = C_{n}^{m-1} \times \frac{n-m+1}{m}$$

利用这个递推公式即可以把空间复杂度优化到 O(k)

代码

package leetcode

func getRow(rowIndex int) []int {
	row := make([]int, rowIndex+1)
	row[0] = 1
	for i := 1; i <= rowIndex; i++ {
		row[i] = row[i-1] * (rowIndex - i + 1) / i
	}
	return row
}

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