Question Link:

https://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.

[-] Python code accepted by LeetCode OJ 
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]