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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?
Solution:
public class Solution { public List<Integer> getRow(int rowIndex) { List<Integer> preRow = new ArrayList<Integer>(); List<Integer> curRow = new ArrayList<Integer>(); curRow.add(1); int len; for (int i=1;i<=rowIndex;i++){ preRow = curRow; curRow = new ArrayList<Integer>(); len = i+1; //j==0 curRow.add(1); //j==1 to (len-2) for (int j=1;j<len-1;j++) curRow.add(preRow.get(j-1)+preRow.get(j)); //j==len-1 curRow.add(1); } return curRow; }}
递推问题。只需存当前行和上一行就行。
Leetcode-Pascal's Triangle II
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