[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?

Pascal's Triangle的扩展题~ 要求返回Pascal's Triangle的第k行，并且只用O(k)的额外空间~ 我们可以从后往前扫，每次需要的数据就是res[j - 1] + res[j]，我们需要的数据不会被覆盖，因为需要的res[j]只在当前步用，下一步就不需要了。

class Solution:
# @return a list of integers
def getRow(self, rowIndex):
if rowIndex < 0: return []
res = [1]
for i in xrange(1, rowIndex + 1):
for j in reversed(xrange(1, i)):
res[j] = res[j - 1] + res[j]
res.append(1)
return res