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"question": {
"questionId": "566",
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"categoryTitle": "Algorithms",
"boundTopicId": 1293,
"title": "Reshape the Matrix",
"titleSlug": "reshape-the-matrix",
"content": "<p>In MATLAB, there is a handy function called <code>reshape</code> which can reshape an <code>m x n</code> matrix into a new one with a different size <code>r x c</code> keeping its original data.</p>\n\n<p>You are given an <code>m x n</code> matrix <code>mat</code> and two integers <code>r</code> and <code>c</code> representing the number of rows and the number of columns of the wanted reshaped matrix.</p>\n\n<p>The reshaped matrix should be filled with all the elements of the original matrix in the same row-traversing order as they were.</p>\n\n<p>If the <code>reshape</code> operation with given parameters is possible and legal, output the new reshaped matrix; Otherwise, output the original matrix.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2021/04/24/reshape1-grid.jpg\" style=\"width: 613px; height: 173px;\" />\n<pre>\n<strong>Input:</strong> mat = [[1,2],[3,4]], r = 1, c = 4\n<strong>Output:</strong> [[1,2,3,4]]\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2021/04/24/reshape2-grid.jpg\" style=\"width: 453px; height: 173px;\" />\n<pre>\n<strong>Input:</strong> mat = [[1,2],[3,4]], r = 2, c = 4\n<strong>Output:</strong> [[1,2],[3,4]]\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>m == mat.length</code></li>\n\t<li><code>n == mat[i].length</code></li>\n\t<li><code>1 &lt;= m, n &lt;= 100</code></li>\n\t<li><code>-1000 &lt;= mat[i][j] &lt;= 1000</code></li>\n\t<li><code>1 &lt;= r, c &lt;= 300</code></li>\n</ul>\n",
"translatedTitle": "重塑矩阵",
"translatedContent": "<p>在 MATLAB 中,有一个非常有用的函数 <code>reshape</code> ,它可以将一个&nbsp;<code>m x n</code> 矩阵重塑为另一个大小不同(<code>r x c</code>)的新矩阵,但保留其原始数据。</p>\n\n<p>给你一个由二维数组 <code>mat</code> 表示的&nbsp;<code>m x n</code> 矩阵,以及两个正整数 <code>r</code> 和 <code>c</code> ,分别表示想要的重构的矩阵的行数和列数。</p>\n\n<p>重构后的矩阵需要将原始矩阵的所有元素以相同的<strong> 行遍历顺序 </strong>填充。</p>\n\n<p>如果具有给定参数的 <code>reshape</code> 操作是可行且合理的,则输出新的重塑矩阵;否则,输出原始矩阵。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2021/04/24/reshape1-grid.jpg\" style=\"width: 613px; height: 173px;\" />\n<pre>\n<strong>输入:</strong>mat = [[1,2],[3,4]], r = 1, c = 4\n<strong>输出:</strong>[[1,2,3,4]]\n</pre>\n\n<p><strong>示例 2</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2021/04/24/reshape2-grid.jpg\" style=\"width: 453px; height: 173px;\" />\n<pre>\n<strong>输入:</strong>mat = [[1,2],[3,4]], r = 2, c = 4\n<strong>输出:</strong>[[1,2],[3,4]]\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示:</strong></p>\n\n<ul>\n\t<li><code>m == mat.length</code></li>\n\t<li><code>n == mat[i].length</code></li>\n\t<li><code>1 &lt;= m, n &lt;= 100</code></li>\n\t<li><code>-1000 &lt;= mat[i][j] &lt;= 1000</code></li>\n\t<li><code>1 &lt;= r, c &lt;= 300</code></li>\n</ul>\n",
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"code": "class Solution {\npublic:\n vector<vector<int>> matrixReshape(vector<vector<int>>& mat, int r, int c) {\n\n }\n};",
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"code": "class Solution {\n public int[][] matrixReshape(int[][] mat, int r, int c) {\n\n }\n}",
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"code": "class Solution(object):\n def matrixReshape(self, mat, r, c):\n \"\"\"\n :type mat: List[List[int]]\n :type r: int\n :type c: int\n :rtype: List[List[int]]\n \"\"\"",
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"code": "public class Solution {\n public int[][] MatrixReshape(int[][] mat, int r, int c) {\n\n }\n}",
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"code": "/**\n * @param {number[][]} mat\n * @param {number} r\n * @param {number} c\n * @return {number[][]}\n */\nvar matrixReshape = function(mat, r, c) {\n\n};",
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"code": "class Solution {\n\n /**\n * @param Integer[][] $mat\n * @param Integer $r\n * @param Integer $c\n * @return Integer[][]\n */\n function matrixReshape($mat, $r, $c) {\n\n }\n}",
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"code": "class Solution {\n List<List<int>> matrixReshape(List<List<int>> mat, int r, int c) {\n \n }\n}",
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"code": "# @param {Integer[][]} mat\n# @param {Integer} r\n# @param {Integer} c\n# @return {Integer[][]}\ndef matrix_reshape(mat, r, c)\n\nend",
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"Do you know how 2d matrix is stored in 1d memory? Try to map 2-dimensions into one.",
"M[i][j]=M[n*i+j] , where n is the number of cols. \r\nThis is the one way of converting 2-d indices into one 1-d index. \r\nNow, how will you convert 1-d index into 2-d indices?",
"Try to use division and modulus to convert 1-d index into 2-d indices.",
"M[i] => M[i/n][i%n] Will it result in right mapping? Take some example and check this formula."
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