【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?

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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：

public class Solution {
    public List<Integer> getRow(int rowIndex) {
        List<Integer> answer = new ArrayList<Integer>();
        answer.add(1);
        
        for(int i = 1;i <= rowIndex;i++){
            List<Integer> temp = new ArrayList<Integer>();
            temp.add(1);
            for(int j = 1;j <= i-1;j++){
                temp.add(answer.get(j-1)+answer.get(j));
            }
            temp.add(1);
            answer = temp;
        }
        
        return answer;
    }
}