566. Reshape the Matrix

转载自：http://algorithm.books.mafengshe.com/answer/566.html

566. Reshape the Matrix

题目描述：

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

题目翻译

在MATLAB中，有一个非常有用的函数叫做“reshape”，它可以将矩阵重新形成一个不同大小但保持其原始数据的新矩阵。

给出一个由二维数组表示的矩阵，两个正整数r和c分别表示所需重组矩阵的行号和列号。

重新组合的矩阵需要以与它们相同的行遍历顺序填充原始矩阵的所有元素。

如果具有给定参数的“重塑”操作是可行且合法的，则输出新的重构矩阵; 否则，输出原始矩阵。

示例1：

输入： 
nums = 
[[1,2]， 
[3,4] 
r = 1，c = 4 
输出： 
[[1,2,3,4]] 
说明：num 
的行遍历是[1,2,3,4]。新的重构矩阵是一个1 * 4矩阵，使用上一个列表逐行填充。 

示例2：

输入： 
nums = 
[[1,2]， 
[3,4] 
r = 1，c = 4 
输出： 
[[1,2,3,4]] 
说明：num 
的行遍历是[1,2,3,4]。新的重构矩阵是一个1 * 4矩阵，使用上一个列表逐行填充。 

注意：
1. 给定矩阵的高度和宽度在[1,100]范围内。
2. 给定的r和c都是正的。

解题方案

标签： Array

思路：将矩阵横向展开为一维数组，元素个数为n=origin_r * origin_c
在元素在一维数组中对应的位置i： 原矩阵位置[i/origin_c,i%origin_c]， 新矩阵位置[i/c,i%c]

代码：

class Solution {
    public int[][] matrixReshape(int[][] nums, int r, int c) {
        int original_r = nums.length;
        int original_c = nums[0].length;
        if (original_r * original_c == r * c) {
            int[][] result = new int[r][c];
            for (int i = 0; i < r * c; i++) {
                result[i / c][i % c] = nums[i / original_c][i % original_c];
            }
            return result;
        } else {
            return nums;
        }
    }
}