Leet Code OJ 119. Pascal's Triangle II [Difficulty: Easy]

题目：
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?

翻译：
给定一个下标k，返回第k行的杨辉三角。
例如给定k=3，返回[1,3,3,1]。
提示：你可以优化你的算法，让它只使用O（k）的额外空间吗？

分析：
如果采用Leet Code OJ 118. Pascal’s Triangle中的方案来做这道题，也就是使用2个Integer数组进行存储，空间复杂度是O(2k)，实际是等价于O（k）的，但是我们考虑一下能否再优化一些。下面的方案去掉lastLine这个数组，采用2个整数代替，空间复杂度为O（k+2），略优于之前方案。

Java版代码:

public class Solution {public List<Integer> getRow(int rowIndex) {List<Integer> line=new ArrayList<>();line.add(1);if(rowIndex==0){return line;}for(int i=1;i<=rowIndex;i++){int lastNum=1;int currentNum=1;for(int j=1;j<i;j++){currentNum=line.get(j);line.set(j,lastNum+currentNum);lastNum=currentNum;}line.add(1);}return line;}
}

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