Leetcode: Pascal's Triangle II

最新推荐文章于 2019-01-11 06:05:08 发布

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最新推荐文章于 2019-01-11 06:05:08 发布

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原文链接：http://www.cnblogs.com/caijinlong/archive/2013/04/30/3051653.html

本文介绍了一个算法问题：如何求出帕斯卡三角形的第K行，并提供了一种解决方案，该方案仅使用O(k)的额外空间。

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?

 1 class Solution {
 2 public:
 3     vector<int> getRow(int rowIndex) {
 4         vector<int> res;
 5         if(rowIndex < 0)
 6             return res;
 7         res.push_back(1);
 8         if(rowIndex == 0)
 9             return res;
10         res.push_back(1);
11         if(rowIndex == 1)
12             return res;
13         int tmp1 = 0;
14         int tmp2 = 0;
15         for(int i = 2; i <= rowIndex; i++)
16         {
17             tmp2 = 1;
18             for(int j = 1; j < i; j++)
19             {
20                 tmp1 = tmp2;
21                 tmp2 = res[j];
22                 res[j] = tmp1+tmp2;
23             }
24             res.push_back(1);
25         }
26         return res;
27     }
28 };