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存量题库数据更新
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@@ -12,29 +12,29 @@
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"translatedContent": "<p>请你编写一段代码为所有数组实现 <code>snail(rowsCount，colsCount)</code> 方法，该方法将 1D 数组转换为以蜗牛排序的模式的 2D 数组。无效的输入值应该输出一个空数组。当 <code>rowsCount * colsCount !==</code><code>nums.length</code> 时。这个输入被认为是无效的。</p>\n\n<p>蜗牛排序从左上角的单元格开始，从当前数组的第一个值开始。然后，它从上到下遍历第一列，接着移动到右边的下一列，并从下到上遍历它。将这种模式持续下去，每列交替变换遍历方向，直到覆盖整个数组。例如，当给定输入数组 <code>[19, 10, 3, 7, 9, 8, 5, 2, 1, 17, 16, 14, 12, 18, 6, 13, 11, 20, 4, 15]</code> ，当 <code>rowsCount = 5</code> 且 <code>colsCount = 4</code> 时，需要输出矩阵如下图所示。注意，矩阵沿箭头方向对应于原数组中数字的顺序</p>\n\n<p> </p>\n\n<p><img alt=\"Traversal Diagram\" src=\"https://assets.leetcode.com/uploads/2023/04/10/screen-shot-2023-04-10-at-100006-pm.png\" style=\"width: 275px; height: 343px;\" /></p>\n\n<p> </p>\n\n<p><b>示例 1：</b></p>\n\n<pre>\n<b>输入：</b>\nnums = [19, 10, 3, 7, 9, 8, 5, 2, 1, 17, 16, 14, 12, 18, 6, 13, 11, 20, 4, 15]\nrowsCount = 5\ncolsCount = 4\n<b>输出：</b>\n[\n [19,17,16,15],\n [10,1,14,4],\n [3,2,12,20],\n [7,5,18,11],\n [9,8,6,13]\n]\n</pre>\n\n<p><b>示例 2：</b></p>\n\n<pre>\n<b>输入：</b>\nnums = [1,2,3,4]\nrowsCount = 1\ncolsCount = 4\n<b>输出：</b>[[1, 2, 3, 4]]\n</pre>\n\n<p><b>示例 3：</b></p>\n\n<pre>\n<b>输入：</b>\nnums = [1,3]\nrowsCount = 2\ncolsCount = 2\n<b>输出：</b>[]\n<strong>Explanation:</strong> 2 * 2 = 4, 且原数组 [1,3] 的长度为 2; 所以，输入是无效的。\n</pre>\n\n<p> </p>\n\n<p><b>提示：</b></p>\n\n<ul>\n\t<li><code>0 <= nums.length <= 250</code></li>\n\t<li><code>1 <= nums[i] <= 1000</code></li>\n\t<li><code>1 <= rowsCount <= 250</code></li>\n\t<li><code>1 <= colsCount <= 250</code></li>\n</ul>\n",
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"lang": "JavaScript",
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"langSlug": "javascript",
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"code": "/**\n * @param {number} rowsCount\n * @param {number} colsCount\n * @return {Array<Array<number>>}\n */\nArray.prototype.snail = function(rowsCount, colsCount) {\n\n}\n\n/**\n * const arr = [1,2,3,4];\n * arr.snail(1,4); // [[1,2,3,4]]\n */",
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"code": "/**\n * @param {number} rowsCount\n * @param {number} colsCount\n * @return {Array<Array<number>>}\n */\nArray.prototype.snail = function(rowsCount, colsCount) {\n \n}\n\n/**\n * const arr = [1,2,3,4];\n * arr.snail(1,4); // [[1,2,3,4]]\n */",
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"lang": "TypeScript",
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"code": "declare global {\n interface Array<T> {\n snail(rowsCount: number, colsCount: number): number[][];\n }\n}\n\nArray.prototype.snail = function(rowsCount: number, colsCount: number): number[][] {\n\n}\n\n/**\n * const arr = [1,2,3,4];\n * arr.snail(1,4); // [[1,2,3,4]]\n */",
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"code": "declare global {\n interface Array<T> {\n snail(rowsCount: number, colsCount: number): number[][];\n }\n}\n\nArray.prototype.snail = function(rowsCount: number, colsCount: number): number[][] {\n \n}\n\n/**\n * const arr = [1,2,3,4];\n * arr.snail(1,4); // [[1,2,3,4]]\n */",
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"hints": [
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"Different ways to approach this problem. Perhaps store a boolean if you are moving up or down and a current column. Reverse the direction and increment the column every time you hits a wall.",
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"Is there a way way to do this without storing state - by just using math?"
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