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?
题解：简单的模拟题，每次用answer列表存放上一层的值，用temp列表存放当前层的值，只要计算好temp中需要重新计算的元素的索引范围[1,i-1](第i层)，然后根据answer计算就可以了，每次计算完一层更新answer为这一层的数，最后answer中存放的就是答案。
代码如下：
1 public class Solution { 2 public List<Integer> getRow(int rowIndex) { 3 List<Integer> answer = new ArrayList<Integer>(); 4 answer.add(1); 5 6 for(int i = 1;i <= rowIndex;i++){ 7 List<Integer> temp = new ArrayList<Integer>(); 8 temp.add(1); 9 for(int j = 1;j <= i-1;j++){ 10 temp.add(answer.get(j-1)+answer.get(j)); 11 } 12 temp.add(1); 13 answer = temp; 14 } 15 16 return answer; 17 } 18 }
【leetcode刷题笔记】Pascal's Triangle II,布布扣,bubuko.com
【leetcode刷题笔记】Pascal's Triangle II
原文：http://www.cnblogs.com/sunshineatnoon/p/3853547.html