119. Pascal's Triangle II [easy] (Python)
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题目链接
https://leetcode.com/problems/pascals-triangle-ii/
题目原文
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?
题目翻译
给定行数k,返回帕斯卡三角形的第k行。比如给定k=3,返回[1,3,3,1]。
备注:你能仅用O(k)的额外空间实现你的算法吗?
思路方法
如果你不太了解帕斯卡三角形,可先参考这一题:
118. Pascal’s Triangle [easy] (Python)
思路一
用到一个小小的trick,即:每一行的结果可以由上一行和上一行的偏移相加得到。例如求第4行:
1 3 3 1 0 + 0 1 3 3 1 = 1 4 6 4 1
代码
class Solution(object): def getRow(self, rowIndex): """ :type rowIndex: int :rtype: List[int] """ res = [1] + [0] * rowIndex for i in range(rowIndex): res[0] = 1 for j in range(i+1, 0, -1): res[j] = res[j] + res[j-1] return res
思路二
另外,也可以不考虑上一行,直接根据公式求出当前行的内容。帕斯卡三角形的第n行第m个元素是 C(n, m) = n!/(m! * (n-m)!)。所以 C(n, m-1) = n!/((m-1)! * (n-m+1)!),所以 C(n, m) = C(n, m-1) * (n-m+1) / m。另外,有组合数的性质可以减少一半计算量:C(n, m) == C(n, n-m)。
代码
class Solution(object): def getRow(self, rowIndex): """ :type rowIndex: int :rtype: List[int] """ res = [1] * (rowIndex+1) for i in range(1, rowIndex/2+1): res[i] = res[rowIndex-i] = res[i-1] * (rowIndex-i+1) / i return res
PS: 新手刷LeetCode,新手写博客,写错了或者写的不清楚还请帮忙指出,谢谢!
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