title: Pascals Triangle II
tags:
- pascals-triangle-ii
- No.119
- simple
- discrete-mathematics
- overflow

Description

Given a non-negative index k where k ≤ 33, return the kth index row of the Pascal's triangle.

Note that the row index starts from 0.

img

In Pascal's triangle, each number is the sum of the two numbers directly above it.

Example:

Input: 3
Output: [1,3,3,1]

Follow up:

Could you optimize your algorithm to use only O(k) extra space?

Corner Cases

• 0
• huge number 30

Solutions

Naive

Use the recursive relationship bellow:

\binom{n}{k+1} = \binom{n}{k} \times \frac{n-k}{k+1} \in \mathbb{N}

Use type Long since the integer overflows when n is big.

The time complexity and space complexity are all O(n):

class Solution {
    public List<Integer> getRow(int rowIndex) {
        List<Integer> lst = new LinkedList<>();
        if (rowIndex == 0) {lst.add(1);return lst;}

        long   cni = 1;
        lst.add((int)cni);
        for (int i=0; i<rowIndex; i++) {
            cni = cni*(rowIndex-i)/(i+1);
            lst.add((int)cni);
        }
        
        return lst;
    }
}