leetcode-566. Reshape the Matrix

https://leetcode.com/problems/reshape-the-matrix/#/description
一道简单的题目，重塑矩阵，按照输入的行数和列数改造矩阵，但是新改造的顺序需要和原矩阵按行遍历的顺序相同。需要判断改造的矩阵是否合理。

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

解题思路：新建一个矩阵，设立的指针自增，并可以控制换行，将原矩阵的数据导入即可，判断一下改造矩阵的格式是否符合规范。

代码如下：

public class Solution {    public int[][] matrixReshape(int[][] nums, int r, int c) {        int row=nums.length;        int column=nums[0].length;        if(row*column!=r*c)        {            return nums;        }        int[][] newArray=new int[r][c];        int m=0,n=0;        for(int i=0;i<row;i++)        {            for(int j=0;j<column;j++)            {                newArray[m][n]=nums[i][j];                n++;                if(n==c)                {                    n=0;                    m++;                }            }        }        return newArray;    }}