Problème
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.
Contraintes
0 ≤ rows ≤ 60. Every entry fits a signed 64-bit integer. Zero rows returns an empty list.
Exemples
Entrée :
rows = 4
Sortie attendue :
expected = [
[1],
[1, 1],
[1, 2, 1],
[1, 3, 3, 1]
]Entrée :
long long rows = 4;
Sortie attendue :
std::vector<std::vector<long long>> expected = {
{1},
{1, 1},
{1, 2, 1},
{1, 3, 3, 1}
};Entrée :
long rows = 4;
Sortie attendue :
long[][] expected = new long[][] {
new long[] {1},
new long[] {1, 1},
new long[] {1, 2, 1},
new long[] {1, 3, 3, 1}
};Explication
The 2 in row 3 is 1 + 1. The two interior 3s in row 4 are 1 + 2 and 2 + 1.
Entrée :
rows = 0
Sortie attendue :
expected = []Entrée :
long long rows = 0;
Sortie attendue :
std::vector<std::vector<long long>> expected = {};Entrée :
long rows = 0;
Sortie attendue :
long[][] expected = new long[][] {};Explication
No rows are requested, so return an empty list.
Comment l'aborder
Essayez d'abord seul, puis ouvrez les indices un par un.
Indice 1
Each row starts and ends with 1, and interior values come from the two values above them.
Indice 2
Build rows one at a time from the previous row rather than recomputing from scratch.
Indice 3
Handle the zero-row case by returning an empty list immediately.
Complexité attendue
O(rows^2) time, O(rows^2) extra space
Références
Ce défi a été adapté pour CodeUtility. Suivez l'énoncé et les contraintes ci-dessus.