题目:
Given a non-negative index k where k ≤ 33, return the kth index row of the Pascal's triangle.
Note that the row index starts from 0.

In Pascal's triangle, each number is the sum of the two numbers directly above it.
Example:
Input: 3 Output: [1,3,3,1]
Follow up:
Could you optimize your algorithm to use only O(k) extra space?
代码:
class Solution {
public:
vector<int> getRow(int rowIndex) {
vector<int> res;
for(int i = 0; i<rowIndex+1; i++)
{
vector<int> cur;
for(int j = 0; j<i+1; j++)
{
if(j == 0 || j == i)
cur.push_back(1);
else
cur.push_back(res[j-1]+res[j]);
}
res = cur;
}
return res;
}
};
加油吧~ 新的一个月,不管结果怎么样,坦然接受吧,加油!
本文介绍了一种解决帕斯卡三角形中特定行元素的算法。通过迭代方式,该算法能有效计算出帕斯卡三角形中任意一行的数值,示例中输入为3,输出为[1,3,3,1]。同时,文章探讨了如何优化算法以减少额外空间使用。
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