What is the key concept behind reshaping a matrix in this problem?

The key concept is converting the matrix from a 2D structure into a 1D array, and then mapping this 1D array back into the new reshaped matrix while maintaining the original row-major order.

How can I handle the case when reshaping is not possible?

If reshaping is not possible (i.e., the total number of elements does not match the product of the desired dimensions), you should return the original matrix.

What is the time complexity of reshaping the matrix?

The time complexity is O(m * n), where m is the number of rows and n is the number of columns in the matrix, due to the need to traverse all elements for reshaping.

Can the matrix be reshaped in-place?

No, reshaping the matrix requires creating a new matrix since the dimensions change. In-place reshaping is not possible for this problem.

What is the role of the primary pattern 'Array plus Matrix' in this problem?

The 'Array plus Matrix' pattern helps in understanding how to flatten a 2D matrix into a 1D array and then rebuild it into a new matrix, ensuring proper data order and efficient memory use.

#566

Easy

auto_awesome数组·matrix

LeetCode 题解工作台

重塑矩阵

在 MATLAB 中，有一个非常有用的函数 reshape ，它可以将一个 m x n 矩阵重塑为另一个大小不同（ r x c ）的新矩阵，但保留其原始数据。给你一个由二维数组 mat 表示的 m x n 矩阵，以及两个正整数 r 和 c ，分别表示想要的重构的矩阵的行数和列数。重构后的矩阵需要…

数组矩阵模拟

题目描述

在 MATLAB 中，有一个非常有用的函数 reshape ，它可以将一个 m x n 矩阵重塑为另一个大小不同（r x c）的新矩阵，但保留其原始数据。

给你一个由二维数组 mat 表示的 m x n 矩阵，以及两个正整数 r 和 c ，分别表示想要的重构的矩阵的行数和列数。

重构后的矩阵需要将原始矩阵的所有元素以相同的 行遍历顺序 填充。

如果具有给定参数的 reshape 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。

示例 1：

输入：mat = [[1,2],[3,4]], r = 1, c = 4
输出：[[1,2,3,4]]

示例 2：

输入：mat = [[1,2],[3,4]], r = 2, c = 4
输出：[[1,2],[3,4]]

提示：

m == mat.length
n == mat[i].length
1 <= m, n <= 100
-1000 <= mat[i][j] <= 1000
1 <= r, c <= 300

lightbulb

解题思路

方法一：模拟

我们先获取原矩阵的行数和列数，分别记为 $m$ 和 $n$ 。如果 $m \times n \neq r \times c$ ，则无法重塑矩阵，直接返回原矩阵。

否则，我们创建一个新矩阵，新矩阵的行数为 $r$ ，列数为 $c$ 。我们从原矩阵的第一个元素开始，按照行优先的顺序遍历原矩阵的所有元素，将遍历到的元素按顺序放入新矩阵中。

遍历完原矩阵的所有元素后，我们即可得到答案。

时间复杂度 $O(m \times n)$ ，其中 $m$ 和 $n$ 分别是原矩阵的行数和列数。忽略答案的空间消耗，空间复杂度 $O(1)$ 。

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class Solution:
    def matrixReshape(self, mat: List[List[int]], r: int, c: int) -> List[List[int]]:
        m, n = len(mat), len(mat[0])
        if m * n != r * c:
            return mat
        ans = [[0] * c for _ in range(r)]
        for i in range(m * n):
            ans[i // c][i % c] = mat[i // n][i % n]
        return ans

speed

复杂度分析

指标	值
时间	Depends on the final approach
空间	Depends on the final approach

psychology

面试官常问的追问

外企场景

question_mark
The candidate should recognize the importance of handling array indexing correctly when reshaping the matrix.
question_mark
Look for the candidate's ability to validate the reshaping dimensions before proceeding.
question_mark
Expect the candidate to explain the time and space complexity of their approach.

warning

常见陷阱

外企场景

error
Failing to check if the reshape dimensions are valid before attempting to reshape.
error
Incorrectly mapping the 1D array back into a 2D matrix, potentially losing elements.
error
Not considering edge cases where the reshape operation is impossible.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Given a 2D matrix, reshape it to match any specified dimensions, with varying boundary conditions (e.g., non-matching dimensions).
arrow_right_alt
Reshape a matrix and also transpose it simultaneously.
arrow_right_alt
Implement an in-place reshape operation that modifies the original matrix.

help

常见问题

外企场景

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