Problem Definition:
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:
1 def getRow(rowIndex): 2 pt=[] 3 for rn in range(rowIndex+1): 4 row=[0]*(rn+1) 5 row[0],row[rn]=1,1 6 for cl in range(1,rn): 7 row[cl]=pt[cl-1]+pt[cl] 8 pt=row 9 return pt
LeetCode#119 Pascal's Triangle II
原文：http://www.cnblogs.com/acetseng/p/4665506.html