Problème
The infinite spiral starts at the top-left cell. The first four rows and columns are shown below. Continue the same pattern in expanding square layers. Coordinates start at 1.
1 2 9 10
4 3 8 11
5 6 7 12
16 15 14 13Return one cell value without constructing the entire grid.
Implement number_spiral(row, column).
Input
row: The row counted from the top, starting at 1.
column: The column counted from the left, starting at 1.
Output
Return the number at that position.
Contraintes
1 ≤ row, column ≤ 1,000,000,000. The value fits a signed 64-bit integer.
Exemples
Entrée :
row = 3
column = 2
Sortie attendue :
expected = 6Entrée :
long long row = 3;
long long column = 2;
Sortie attendue :
long long expected = 6;Entrée :
long row = 3;
long column = 2;
Sortie attendue :
long expected = 6;Explication
In the displayed third row, the second value is 6.
Entrée :
row = 1
column = 4
Sortie attendue :
expected = 10Entrée :
long long row = 1;
long long column = 4;
Sortie attendue :
long long expected = 10;Entrée :
long row = 1;
long column = 4;
Sortie attendue :
long expected = 10;Explication
The fourth square layer begins at 10 in the top row.
Comment l'aborder
Essayez d'abord seul, puis ouvrez les indices un par un.
Indice 1
The spiral is built in square layers, so a cell belongs to a layer determined by the larger coordinate.
Indice 2
Use the parity of that layer to decide whether values increase along a row or a column.
Indice 3
Pay attention to whether the layer reaches from the top-left corner or the opposite corner first.
Complexité attendue
O(1) time, O(1) extra space
Références
Ce défi a été adapté pour CodeUtility. Suivez l'énoncé et les contraintes ci-dessus.