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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?
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解题分析：
此处有空间的限制，因此不能正常使用迭代算法，
这里推荐滚动数组思想，具体思想在我的上一篇博客里有写。
# -*- coding:utf-8 -*-
__author__ = 'jiuzhang'
class Solution(object):
def getRow(self, rowIndex):
if rowIndex < 0:
return []
result = [0] * (rowIndex + 1)
for i in range(rowIndex + 1):
result[i] = 1
for j in xrange(i - 1, 0, -1):
result[j] += result[j - 1]
return result
(LeetCode)Pascal's Triangle II --- 杨辉三角进阶（滚动数组思想）
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原文地址：http://blog.csdn.net/u012965373/article/details/52180940
