leetcode | Pascal's Triangle II

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?

第n行的第r个元素的表达式是nCr, 就是排列组合，有一个公式  nCr = (n-r+1)/r  *  nC(r-1);

public class Solution {    public List<Integer> getRow(int rowIndex) {            List<Integer> row = new ArrayList<>();            int n = rowIndex+1;            for(int i=0;i<n;i++)            {                row.add((int)C(rowIndex,i));            }            return row;    }                                public long C(int n,int r)        {            long re = 1;            for(int i=1;i<=r;i++)            {                re = re*(n-i+1)/i;            }            return re;        }}