[LeetCode] 566. Reshape the Matrix

最新推荐文章于 2022-10-31 17:32:30 发布

锤子剪子布

最新推荐文章于 2022-10-31 17:32:30 发布

阅读量169

点赞数

CC 4.0 BY-SA版权

分类专栏： LeetCode 文章标签： LeetCode

本文链接：https://blog.youkuaiyun.com/mirinta/article/details/81194441

LeetCode 专栏收录该内容

30 篇文章

订阅专栏

本文介绍了一个类似于MATLAB中'reshape'函数的算法实现，该算法接收一个二维数组和两个整数作为参数，返回一个新的矩阵，其行数和列数分别由这两个整数决定。如果输入的参数不合法，即新矩阵的元素数量与原矩阵不符，算法将返回原矩阵。通过实例演示了如何使用该算法。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

In MATLAB, there is a very useful function called 'reshape', which can reshape a matrix into a new one with different size but keep its original data.

You're given a matrix represented by a two-dimensional array, and two positive integers r and c representing the row number and column number of the wanted reshaped matrix, respectively.

The reshaped matrix need to be filled with all the elements of the original matrix in the same row-traversing order as they were.

If the 'reshape' operation with given parameters is possible and legal, output the new reshaped matrix; Otherwise, output the original matrix.

Example 1:

Input: 
nums = 
[[1,2],
 [3,4]]
r = 1, c = 4
Output: 
[[1,2,3,4]]
Explanation:
The row-traversing of nums is [1,2,3,4]. The new reshaped matrix is a 1 * 4 matrix, fill it row by row by using the previous list.

Example 2:

Input: 
nums = 
[[1,2],
 [3,4]]
r = 2, c = 4
Output: 
[[1,2],
 [3,4]]
Explanation:
There is no way to reshape a 2 * 2 matrix to a 2 * 4 matrix. So output the original matrix.

Note:

The height and width of the given matrix is in range [1, 100].
The given r and c are all positive.

题目：输入一个二维数组和两个正整数r与c，实现MATLAB中矩阵的reshape方法，返回行数为r、列数为n的新矩阵（数组）。如果输入的r和c是非法参数，则返回原矩阵（数组）。

实现思路：假设原矩阵的行数为rows、列数为cols，由于原矩阵与新矩阵必须保证相同的元素数量，因此r和c的合法条件是r * c = rows * cols。以下讨论的矩阵，行和列都从0开始计数：由于reshape按照行优先的顺序进行，可以想象将原矩阵的各行并排放置（拉直成一个只有一行的矩阵），此时原矩阵中第i行第j列的元素就是并排矩阵的第i * cols + j 个元素，令N = i * cols + j。假设这个元素应该放在新矩阵的第p行第q列，则利用并排矩阵的道理可知p * c + q = N，因此有p = N / c，q = N % c。

class Solution {
    public int[][] matrixReshape(int[][] nums, int r, int c) {
        if (nums == null) throw new IllegalArgumentException("argument is null");
        int rows = nums.length, cols = rows == 0 ? 0 : nums[0].length;
        if (r * c != rows * cols) return nums;
        int[][] result = new int[r][c];
        for (int count = 0; count < rows * cols; count++)
            result[count / c][count % c] = nums[count / cols][count % cols];
        return result;
    }
}