Условие
The first row is [1]. Each later row starts and ends with 1. Every interior entry is the sum of the two entries immediately above it. Return the first rows rows, beginning with the single 1.
Implement pascals_triangle(rows).
Input
rows: How many rows to generate.
Output
Return a list of rows, starting at the top; each row is a list of integers.
Ограничения
0 ≤ rows ≤ 60. Every entry fits a signed 64-bit integer. Zero rows returns an empty list.
Примеры
Ввод:
rows = 4
Ожидаемый вывод:
expected = [
[1],
[1, 1],
[1, 2, 1],
[1, 3, 3, 1]
]Ввод:
long long rows = 4;
Ожидаемый вывод:
std::vector<std::vector<long long>> expected = {
{1},
{1, 1},
{1, 2, 1},
{1, 3, 3, 1}
};Ввод:
long rows = 4;
Ожидаемый вывод:
long[][] expected = new long[][] {
new long[] {1},
new long[] {1, 1},
new long[] {1, 2, 1},
new long[] {1, 3, 3, 1}
};Пояснение
The 2 in row 3 is 1 + 1. The two interior 3s in row 4 are 1 + 2 and 2 + 1.
Ввод:
rows = 0
Ожидаемый вывод:
expected = []Ввод:
long long rows = 0;
Ожидаемый вывод:
std::vector<std::vector<long long>> expected = {};Ввод:
long rows = 0;
Ожидаемый вывод:
long[][] expected = new long[][] {};Пояснение
No rows are requested, so return an empty list.
Как подступиться
Сначала попробуйте сами, затем открывайте подсказки по одной.
Подсказка 1
Each row starts and ends with 1, and interior values come from the two values above them.
Подсказка 2
Build rows one at a time from the previous row rather than recomputing from scratch.
Подсказка 3
Handle the zero-row case by returning an empty list immediately.
Ожидаемая сложность
O(rows^2) time, O(rows^2) extra space
Ссылки
Это задание адаптировано для CodeUtility. Следуйте условию и ограничениям выше.