Problem Link:

http://oj.leetcode.com/problems/pascals-triangle-ii/

Let T[i][j] be the j-th element of the i-th row in the triangle, for 0 <= j <= i, i = 0, 1, ...

And we have the recursive function for T[i][j]

T[i][j] = 1, if j == 0 or j == i

T[i][j] = T[i-1][j] + T[i-1][j-1]

Simple enough, the python code is as follows.

class Solution:
# @return a list of integers
def getRow(self, rowIndex):
# Spcecial case
if rowIndex < 2:
return [1] * (rowIndex + 1)
# Initial two lists for update
res = []
res.append([1]*(rowIndex+1))
res.append([1]*(rowIndex+1))
current = 0
# Update row by row
for row in xrange(2, rowIndex+1):
for i in xrange(1, row):
res[1-current][i] = res[current][i] + res[current][i-1]
current = 1 - current
# Return the current list
return res[current]

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