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            "title": "Spiral Matrix",
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            "content": "<p>Given an <code>m x n</code> <code>matrix</code>, return <em>all elements of the</em> <code>matrix</code> <em>in spiral order</em>.</p>\n\n<p>&nbsp;</p>\n<p><strong>Example 1:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2020/11/13/spiral1.jpg\" style=\"width: 242px; height: 242px;\" />\n<pre>\n<strong>Input:</strong> matrix = [[1,2,3],[4,5,6],[7,8,9]]\n<strong>Output:</strong> [1,2,3,6,9,8,7,4,5]\n</pre>\n\n<p><strong>Example 2:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2020/11/13/spiral.jpg\" style=\"width: 322px; height: 242px;\" />\n<pre>\n<strong>Input:</strong> matrix = [[1,2,3,4],[5,6,7,8],[9,10,11,12]]\n<strong>Output:</strong> [1,2,3,4,8,12,11,10,9,5,6,7]\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>m == matrix.length</code></li>\n\t<li><code>n == matrix[i].length</code></li>\n\t<li><code>1 &lt;= m, n &lt;= 10</code></li>\n\t<li><code>-100 &lt;= matrix[i][j] &lt;= 100</code></li>\n</ul>\n",
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                    "code": "class Solution {\npublic:\n    vector<int> spiralOrder(vector<vector<int>>& matrix) {\n        \n    }\n};",
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                    "code": "class Solution {\n    public List<Integer> spiralOrder(int[][] matrix) {\n        \n    }\n}",
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                    "code": "class Solution(object):\n    def spiralOrder(self, matrix):\n        \"\"\"\n        :type matrix: List[List[int]]\n        :rtype: List[int]\n        \"\"\"\n        ",
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                    "lang": "Python3",
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                    "code": "class Solution:\n    def spiralOrder(self, matrix: List[List[int]]) -> List[int]:\n        ",
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                    "lang": "C",
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                    "code": "\n\n/**\n * Note: The returned array must be malloced, assume caller calls free().\n */\nint* spiralOrder(int** matrix, int matrixSize, int* matrixColSize, int* returnSize){\n\n}",
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                    "code": "public class Solution {\n    public IList<int> SpiralOrder(int[][] matrix) {\n        \n    }\n}",
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                    "code": "/**\n * @param {number[][]} matrix\n * @return {number[]}\n */\nvar spiralOrder = function(matrix) {\n    \n};",
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                    "code": "# @param {Integer[][]} matrix\n# @return {Integer[]}\ndef spiral_order(matrix)\n    \nend",
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                    "lang": "Swift",
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                    "code": "class Solution {\n    func spiralOrder(_ matrix: [[Int]]) -> [Int] {\n        \n    }\n}",
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                    "lang": "Go",
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                    "code": "func spiralOrder(matrix [][]int) []int {\n    \n}",
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                    "lang": "Scala",
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                    "code": "object Solution {\n    def spiralOrder(matrix: Array[Array[Int]]): List[Int] = {\n        \n    }\n}",
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                    "lang": "Kotlin",
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                    "code": "class Solution {\n    fun spiralOrder(matrix: Array<IntArray>): List<Int> {\n        \n    }\n}",
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                    "lang": "Rust",
                    "langSlug": "rust",
                    "code": "impl Solution {\n    pub fn spiral_order(matrix: Vec<Vec<i32>>) -> Vec<i32> {\n        \n    }\n}",
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                    "lang": "PHP",
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                    "code": "class Solution {\n\n    /**\n     * @param Integer[][] $matrix\n     * @return Integer[]\n     */\n    function spiralOrder($matrix) {\n        \n    }\n}",
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                    "code": "(define/contract (spiral-order matrix)\n  (-> (listof (listof exact-integer?)) (listof exact-integer?))\n\n  )",
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                    "lang": "Erlang",
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            "hints": [
                "Well for some problems, the best way really is to come up with some algorithms for simulation. Basically, you need to simulate what the problem asks us to do.",
                "We go boundary by boundary and move inwards. That is the essential operation. First row, last column, last row, first column and then we move inwards by 1 and then repeat. That's all, that is all the simulation that we need.",
                "Think about when you want to switch the progress on one of the indexes. If you progress on <pre>i</pre> out of <pre>[i, j]</pre>, you'd be shifting in the same column. Similarly, by changing values for <pre>j</pre>, you'd be shifting in the same row.\r\nAlso, keep track of the end of a boundary so that you can move inwards and then keep repeating. It's always best to run the simulation on edge cases like a single column or a single row to see if anything breaks or not."
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