[Leetcode] Pascal's Triangle II
2016-12-16 13:08
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描述
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?
分析
给定 n 返回杨辉三角的第 n 行。仿照 前一题 [Pascal’s Triangle] 的思路进行构造。由于这里不需要返回前 n 行而只需要第 n 行,因此可以共用一个向量,每次将杨辉三角的下一行赋值给这个向量,直到第 n 行,这样空间复杂度为 0(n) 。
代码
class Solution { public: vector<int> getRow(int rowIndex) { vector<int> res(rowIndex + 1, 0); for (int i = 0; i <= rowIndex; i++) { res[0] = res[i] = 1; for (int j = i - 1; j > 0; j--) res[j] += res[j - 1]; } return res; } };
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