2022. Convert 1 D Array Into 2 D Array

2022. Convert 1D Array Into 2D Array #

题目 #

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:

https://assets.leetcode.com/uploads/2021/08/26/image-20210826114243-1.png

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

题目大意 #

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。

解题思路 #

  • 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。

代码 #

package leetcode

func construct2DArray(original []int, m int, n int) [][]int {
	if m*n != len(original) {
		return [][]int{}
	}
	res := make([][]int, m)
	for i := 0; i < m; i++ {
		res[i] = original[n*i : n*(i+1)]
	}
	return res
}

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