LeetCode题解:Pascal's Triangle II
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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; }}
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