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77 lines
2.8 KiB
Markdown
77 lines
2.8 KiB
Markdown
# [2022. Convert 1D Array Into 2D Array](https://leetcode.com/problems/convert-1d-array-into-2d-array/)
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## 题目
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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`.
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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.
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Return *an* `m x n` *2D array constructed according to the above procedure, or an empty 2D array if it is impossible*.
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**Example 1:**
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```
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Input: original = [1,2,3,4], m = 2, n = 2
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Output: [[1,2],[3,4]]
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Explanation: The constructed 2D array should contain 2 rows and 2 columns.
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The first group of n=2 elements in original, [1,2], becomes the first row in the constructed 2D array.
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The second group of n=2 elements in original, [3,4], becomes the second row in the constructed 2D array.
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```
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**Example 2:**
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```
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Input: original = [1,2,3], m = 1, n = 3
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Output: [[1,2,3]]
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Explanation: The constructed 2D array should contain 1 row and 3 columns.
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Put all three elements in original into the first row of the constructed 2D array.
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```
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**Example 3:**
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```
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Input: original = [1,2], m = 1, n = 1
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Output: []
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Explanation: There are 2 elements in original.
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It is impossible to fit 2 elements in a 1x1 2D array, so return an empty 2D array.
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```
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**Constraints:**
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- `1 <= original.length <= 5 * 104`
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- `1 <= original[i] <= 105`
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- `1 <= m, n <= 4 * 104`
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## 题目大意
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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。
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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。
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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。
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## 解题思路
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- 简单题。从一维数组 original 中依次取出每行 n 个元素，顺序放到 m 行中。此题中，如果 m*n 大于或者小于 original 的长度，都输出空数组。
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## 代码
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```go
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package leetcode
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func construct2DArray(original []int, m int, n int) [][]int {
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if m*n != len(original) {
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return [][]int{}
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}
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res := make([][]int, m)
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for i := 0; i < m; i++ {
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res[i] = original[n*i : n*(i+1)]
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}
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return res
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}
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``` |