Leetcode 细节实现 Pascal's Triangle II

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Pascal's Triangle II

 Total Accepted: 9401 Total Submissions: 31433

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?

题意：题意：给定数字rowIndex，返回杨辉三角的第rowIndex行
思路：按杨辉三角的定义生成rowIndex+1行杨辉，然后返回第rowIndex行即可
复杂度：时间O(n^2)，空间O(n^2)

class Solution {public:    vector<int> getRow(int rowIndex) {    vector<int > triangle;        int numRows = rowIndex + 1;        if(numRows == 0) return triangle;        vector<int> row1; row1.push_back(1);        triangle = row1;         for(int i = 1; i < numRows; i++){        vector<int> row;        row.push_back(1);        for(int j = 0; j < i - 1; j++){        row.push_back(triangle[j] + triangle[j + 1]);        }        row.push_back(1);        triangle = row;        }        return triangle;    }};