Swift Spiral Matrix - arrays

Question

Question
Given an integer n, generate a square matrix filled with
elements from 1 to n^2 in spiral order.
For example, Given n = 3,
You should return the following matrix: [ [ 1, 2, 3 ], [ 8, 9, 4 ], [
7, 6, 5 ] ]
class Solution
{
func generateMatrix(n: Int) -> [[Int]]
{
var matrix =
[[Int]](count: n, repeatedValue: [Int](count: n, repeatedValue: 0))
var left = 0
var top = 0
var right = n - 1
var down = n - 1
var count = 1
while left <= right && top < down // shouble left <= right && top <= down
{
for j in (left...right)
{
matrix[top][j] = count
count += 1
}
top += 1
for i in (top...down)
{
matrix[i][right] = count
count += 1
}
right -= 1
for j in (left...right).reverse()
{
matrix[down][j] = count
count += 1
}
down -= 1
for i in (top...down).reverse()
{
matrix[i][left] = count
count += 1
}
left += 1
}
return matrix
}
}
var test = Solution()
var result = test.generateMatrix(3)
print(result)
This is my result, [ [ 1, 2, 3 ], [ 8, 0, 4 ], [ 7, 6, 5 ] ], the 9 is
missing. I guess I should change my while loop to "left <= right &&
top <= down",but I got an error.
The error makes me so confusing. Since under the condition "left <=
right && top <= down", variable top has no chance to get over variable
down, however, the error alerts me range end < start.
Thank you so much for you help! Really appreciate your time.

Answer 1

Look like homework but I'll bite. Assuming i, j are the row and column indexes starting at 0, 0, the spiral movement can be described as follow:
Increase j until you hit the matrix boundary on the right
Increase i until you hit the matrix boundary at the bottom
Decrease j until you hit the matrix boundary on the left
Decrease i until you hit the matrix boundary at the top
Repeat step 1 - 4, substituting "matrix boundary" for "non-empty cell"
Here's the code:
let n = 3
// Init an all-zero matrix
var matrix = Array(0..<n).map { _ in [Int](count: n, repeatedValue: 0) }
var i = 0
var j = 0
// These 2 variables control how we move the i and j cursors
// Initial move is from left to right
var deltaI = 0
var deltaJ = 1
for number in 1...(n*n) {
matrix[i][j] = number
let nextI = i + deltaI
let nextJ = j + deltaJ
// nextCellIsEmpty == true if:
// * nextI is within boundary of the matrix; and
// * nextJ is within boundary of the matrix; and
// * matrix[nextI][nextJ] is not taken
let nextCellIsEmpty = (0..<n ~= nextI) && (0..<n ~= nextJ) && (matrix[nextI][nextJ] == 0)
// If the next cell is not empty, we need to adjust how
// the cursors move
if !nextCellIsEmpty {
if deltaJ == 1 { deltaI = 1; deltaJ = 0; }
else if deltaI == 1 { deltaI = 0; deltaJ = -1; }
else if deltaJ == -1 { deltaI = -1; deltaJ = 0; }
else if deltaI == -1 { deltaI = 0; deltaJ = 1; }
}
i += deltaI
j += deltaJ
}
matrix.forEach { print($0) }
~= is the "pattern match" operator. a..<b ~= c returns true if a <= c < b

Answer 2

add this code in the end:
matrix[(bottom + top) / 2][(left + right) / 2] = n * n
because of your algorithm, the matrix in the center can not meet your limit,
so just add this code.
furthermore，the n should be odd number（1,3,5...） that can meet your algorithm,that should be considered.