You are given a 0-indexed 1-dimensional (1D) integer array original, and two integers, m and n. You are tasked with creating a 2-dimensional (2D) array with m rows and n columns using all the elements from original.

The elements from indices 0 to n - 1 (inclusive) of original should form the first row of the constructed 2D array, the elements from indices n to 2 * n - 1 (inclusive) should form the second row of the constructed 2D array, and so on.

Return an m x n 2D array constructed according to the above procedure, or an empty 2D array if it is impossible.

Example 1:

Input: original = [1,2,3,4], m = 2, n = 2
Output: [[1,2],[3,4]]
Explanation: The constructed 2D array should contain 2 rows and 2 columns.
The first group of n=2 elements in original, [1,2], becomes the first row in the constructed 2D array.
The second group of n=2 elements in original, [3,4], becomes the second row in the constructed 2D array.

Example 2:

Input: original = [1,2,3], m = 1, n = 3
Output: [[1,2,3]]
Explanation: The constructed 2D array should contain 1 row and 3 columns.
Put all three elements in original into the first row of the constructed 2D array.

Example 3:

Input: original = [1,2], m = 1, n = 1
Output: []
Explanation: There are 2 elements in original.
It is impossible to fit 2 elements in a 1x1 2D array, so return an empty 2D array.

Constraints:

• 1 <= original.length <= 5 * 104
• 1 <= original[i] <= 105
• 1 <= m, n <= 4 * 104

将一维数组转变成二维数组。

给你一个下标从 0 开始的一维整数数组 original 和两个整数 m 和  n 。你需要使用 original 中 所有 元素创建一个 m 行 n 列的二维数组。

original 中下标从 0 到 n - 1 （都 包含 ）的元素构成二维数组的第一行，下标从 n 到 2 * n - 1 （都 包含 ）的元素构成二维数组的第二行，依此类推。

请你根据上述过程返回一个 m x n 的二维数组。如果无法构成这样的二维数组，请你返回一个空的二维数组。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/convert-1d-array-into-2d-array
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这是一道关于矩阵的模拟题，用心做即可。把一个一维矩阵转换成一个二维矩阵。注意这里的一个 corner case 是如果一维矩阵的长度不等于二维矩阵的面积的话，就返回一个空的矩阵即可。

时间O(n)

空间O(mn)

Java实现

 1 class Solution {
 2     public int[][] construct2DArray(int[] original, int m, int n) {
 3         // corner case
 4         if (original.length != m * n) {
 5             return new int[0][0];
 6         }
 7 
 8         // normal case
 9         int[][] res = new int[m][n];
10         for (int i = 0; i < original.length; i++) {
11             res[i / n][i % n] = original[i];
12         }
13         return res;
14     }
15 }

 

LeetCode 题目总结