mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
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182 lines
16 KiB
JSON
182 lines
16 KiB
JSON
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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> </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> </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 <= m, n <= 10</code></li>\n\t<li><code>-100 <= matrix[i][j] <= 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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"code": "impl Solution {\n pub fn spiral_order(matrix: Vec<Vec<i32>>) -> Vec<i32> {\n \n }\n}",
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"code": "function spiralOrder(matrix: number[][]): number[] {\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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"code": "-spec spiral_order(Matrix :: [[integer()]]) -> [integer()].\nspiral_order(Matrix) ->\n .",
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"code": "defmodule Solution do\n @spec spiral_order(matrix :: [[integer]]) :: [integer]\n def spiral_order(matrix) do\n\n end\nend",
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"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.",
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"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.",
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"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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