Leetcode-Pascal's Triangle II

Given an index k, return the k^th 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。