LeetCode题解：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：给定k，返回Pascal 三角形的第k行，要求O(k)空间

思路：和Pascal’s Triangle一样，只是不用存所有了……

代码：

public class Solution {
public List<Integer> getRow(int rowIndex) {
List<Integer> row = new ArrayList<Integer>();

for(int i = 0;i < rowIndex + 1;++i){
row.add(0,1);

for(int j = 1;j < row.size() - 1;++j){
row.set(j, row.get(j) + row.get(j + 1));
}
}

return row;
}
}