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