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Given an m x n matrix, return all elements of the matrix in spiral order.
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/** | |
* @param {number[][]} matrix | |
* @return {number[]} | |
* https://underdash.web.app | |
* https://leetcode.com/problems/spiral-matrix/ | |
* https://underdash.tech/spiral-matrix/ | |
* Copyright reserved by Taha Shieenavaz (Underdash) | |
*/ | |
const spiralOrder = matrix => { | |
const { abs } = Math | |
const rows = matrix.length, cols = matrix[0].length | |
const map = Array(rows).fill().map(row => { | |
return Array(cols).fill(false) | |
}) | |
const result = [] | |
const direction = { x: 1, y: 0 } | |
let i = 0, j = 0 | |
while( map.flat().filter( a => !a ).length > 0 ) { | |
if( | |
( direction.x == -1 && i == 0 ) || | |
( direction.y == -1 && j == 0 ) || | |
( direction.x == 1 && i == cols - 1 ) || | |
( direction.y == 1 && j == rows - 1 ) || | |
map[j+direction.y][i+direction.x] | |
){ | |
if( abs(direction.x) ) { | |
direction.y = direction.x | |
direction.x = 0 | |
}else if( abs(direction.y) ) { | |
direction.x = -1 * direction.y | |
direction.y = 0 | |
} | |
} | |
result.push(matrix[j][i]) | |
map[j][i] = true | |
i += direction.x | |
j += direction.y | |
} | |
return result | |
}; | |
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Read more about this solution: https://underdash.pro/spiral-matrix/