119. Pascal's Triangle II

题目

解题思路

此题和117杨辉三角的思路一样。循环所有由两个数字组成的index，最后即可得到结果。
pascal[i][j] = pascal[i-1][j-1] + pascal[i-1][j]

class Solution:
    def getRow(self, rowIndex: int) -> List[int]:
        pascal = [[1] * (i+1) for i in range(rowIndex+1)]
        for i in range(rowIndex+1):
            for j in range(1,i):
                pascal[i][j] = pascal[i-1][j-1] + pascal[i-1][j]
        return pascal[-1]