【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：

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;
}
}