LeetCode 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?

题意：输入k行，输出第k行的杨辉三角，有空间限制，只能有O(k)

思路：第k行实际上是在第k-1行的结果上追加1，然后从倒数第二个开始，就等于它与前一个的和

public class Solution {public List<Integer> getRow(int rowIndex) {ArrayList<Integer> list = new ArrayList<Integer>();if (rowIndex == 0) {list.add(1);} else {list.add(1);for (int i = 2; i <= rowIndex + 1; i++) {list.add(1);for (int len = list.size(), j = len - 2; j >= 1; j--) {int tmp = list.get(j) + list.get(j - 1);list.set(j, tmp);}}}return list;}
}

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