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【leetcode刷题笔记】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?


 

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 }