「这是我参与11月更文挑战的第7天，活动详情查看：2021最后一次更文挑战」
描述
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.
Example 4:
Input: original = [3], m = 1, n = 2
Output: []
Explanation:
There is 1 element in original.
It is impossible to make 1 element fill all the spots in a 1x2 2D array, so return an empty 2D array.
Note:
1 <= original.length <= 5 * 10^4
1 <= original[i] <= 10^5
1 <= m, n <= 4 * 104^
解析
根据题意，就是给出了一个从 0 开始索引的一维列表 original ，并且有两个整数 m 和 n ，我们的任务就是创建一个 m 行 n 列的二维列表，值都是用的 original 中的值，原始索引 0 到 n - 1（含）的元素应构成构造的二维数组的第一行，索引 n 到 2 * n - 1（含）的元素应构成构造的二维数组的第二行， 以此类推等等。
返回一个按照上述过程构造的 m x n 二维数组，如果不可能，则返回一个空的二维数组。
题干这么长，其实很简单，就是个纸老虎，一个能打的都没有！思路：
- 先判断 m*n 如果和 original 长度不相等，直接返回空列表
- 否则初始化一个 result ，然后从左到右每次截取 n 个 original 中的元素加到 result 后面
- 遍历结束，返回 result
解答
class Solution(object):
def construct2DArray(self, original, m, n):
"""
:type original: List[int]
:type m: int
:type n: int
:rtype: List[List[int]]
"""
if m*n!=len(original) : return []
result = []
for i in range(m):
result.append(original[i*n:(i+1)*n])
return result
运行结果
Runtime: 884 ms, faster than 89.55% of Python online submissions for Convert 1D Array Into 2D Array.
Memory Usage: 22 MB, less than 50.87% of Python online submissions for Convert 1D Array Into 2D Array.
原题链接：leetcode.com/problems/co…
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