SpiralMatrix.swift

/**
 * Question Link: https://leetcode.com/problems/spiral-matrix/
 * Primary idea: Use four index to get the right element during iteration
 *
 * Time Complexity: O(n^2), Space Complexity: O(1)
 */

class SpiralMatrix {
    func spiralOrder(_ matrix: [[Int]]) -> [Int] {
        var res = [Int]()
    
        guard matrix.count != 0 else {
            return res
        }
    
        var startX = 0
        var endX = matrix.count - 1
        var startY = 0
        var endY = matrix[0].count - 1
        
        while true {
            // top
            for i in startY...endY {
                res.append(matrix[startX][i])
            }
            startX += 1
            if startX > endX {
                break
            }
            
            // right
            for i in startX...endX {
                res.append(matrix[i][endY])
            }
            endY -= 1
            if startY > endY {
                break
            }
            
            // bottom
            for i in stride(from: endY, through: startY, by: -1) {
                res.append(matrix[endX][i])
            }
            endX -= 1
            if startX > endX {
                break
            }
            
            // left
            for i in stride(from: endX, through: startX, by: -1) {
                res.append(matrix[i][startY])
            }
            startY += 1
            if startY > endY {
                break
            }
        }
        
        return res
    }
}