leetcode-problemset/leetcode-cn/originData/reshape-the-matrix.json

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            "title": "Reshape the Matrix",
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            "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        \"\"\"\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": "function matrixReshape(mat: number[][], r: number, c: number): number[][] {\n    \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 * Return an array of arrays of size *returnSize.\n * The sizes of the arrays are returned as *returnColumnSizes array.\n * Note: Both returned array and *columnSizes array must be malloced, assume caller calls free().\n */\nint** matrixReshape(int** mat, int matSize, int* matColSize, int r, int c, int* returnSize, int** returnColumnSizes) {\n    \n}",
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                    "code": "class Solution {\n    fun matrixReshape(mat: Array<IntArray>, r: Int, c: Int): Array<IntArray> {\n        \n    }\n}",
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                    "code": "class Solution {\n    func matrixReshape(_ mat: [[Int]], _ r: Int, _ c: Int) -> [[Int]] {\n        \n    }\n}",
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                    "code": "impl Solution {\n    pub fn matrix_reshape(mat: Vec<Vec<i32>>, r: i32, c: i32) -> Vec<Vec<i32>> {\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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                    "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": "object Solution {\n    def matrixReshape(mat: Array[Array[Int]], r: Int, c: Int): Array[Array[Int]] = {\n        \n    }\n}",
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                    "code": "defmodule Solution do\n  @spec matrix_reshape(mat :: [[integer]], r :: integer, c :: integer) :: [[integer]]\n  def matrix_reshape(mat, r, c) do\n    \n  end\nend",
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                    "code": "-spec matrix_reshape(Mat :: [[integer()]], R :: integer(), C :: integer()) -> [[integer()]].\nmatrix_reshape(Mat, R, C) ->\n  .",
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                    "code": "(define/contract (matrix-reshape mat r c)\n  (-> (listof (listof exact-integer?)) exact-integer? exact-integer? (listof (listof exact-integer?)))\n  )",
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            "hints": [
                "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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