119 Pascal's Triangle II - Easy

Problem:

Given an integer rowIndex, return the rowIndexth (0-indexed) row of the Pascal's triangle.

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

Example 1:

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

Example 2:

Input: rowIndex = 0 Output: [1]

Example 3:

Input: rowIndex = 1 Output: [1,1]

Constraints:

• 0 <= rowIndex <= 33

Follow up: Could you optimize your algorithm to use only O(rowIndex) extra space?

Problem Analysis:

• slight modification from 118-pascals-triangle118 Pascal's Triangle - EasyProblem: Given an integer numRows, return the first numRows of Pascal's triangle. In Pascal's triangle, each number is the sum of the two numbers directly above it as shown: Example 1: Input: numRows = 5 Output: [[1],[1,1],[1,2,1],[1,3,3,1],[1,4,6,4,1]] Example 2: Input: numRows = 1 Output: [[1]] Constraints: * 1 <= numRows <= 30 Problem Analysis: * dynamic programming Trick**: add zero at the side Solutions: class Solution(object): def generate(self, numRows): """

Solutions:

class Solution:
    def getRow(self, rowIndex: int) -> List[int]:
        res = [[1]]

        for i in range(rowIndex):
            # previous row, add 0 at the side for easy calculation
            temp = [0] + res[-1] + [0]
            row = []
            for j in range(len(res[-1])+1):
                row.append(temp[j] + temp[j+1])
            res.append(row)

        return res[-1]      

Similar Questions

118-pascals-triangle118 Pascal's Triangle - EasyProblem: Given an integer numRows, return the first numRows of Pascal's triangle. In Pascal's triangle, each number is the sum of the two numbers directly above it as shown: Example 1: Input: numRows = 5 Output: [[1],[1,1],[1,2,1],[1,3,3,1],[1,4,6,4,1]] Example 2: Input: numRows = 1 Output: [[1]] Constraints: * 1 <= numRows <= 30 Problem Analysis: * dynamic programming Trick**: add zero at the side Solutions: class Solution(object): def generate(self, numRows): """