题目链接：https://leetcode.com/problems/reshape-the-matrix/
原题描述：
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:
Therow-traversingof 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.
题目大意：
（简单题）给一个矩阵，矩阵由若干数组组成，给一对行 与列的值，要求将数组变换成给定行列大小的矩阵，称之为 reshape
解题方法（java语言）：
方法一：将所给数组转为一维数组，在开辟一个二维数组，将一维数组的各个数字存进去（队列的思想）
class Solution {
public int[][] matrixReshape(int[][] nums, int r, int c) {
int row = nums.length;
int col = nums[0].length;
if(row*col!=r*c||nums.length==0){return nums;}
int[] arr1 = new int[row*col];
int k=0;
for(int i=0;i方法二：思想同上，用了队列而已，在这里可以熟悉一下java中队列的操作
| 操作 | 说明 | 备注 |
|---|---|---|
| add | 增加一个元素 | 如果队列已满，则抛出一个IIIegaISlabEepeplian异常 |
| remove | 移除并返回队列头部的元素 | 如果队列为空，则抛出一个NoSuchElementException异常 |
| element | 返回队列头部的元素 | 如果队列为空，则抛出一个NoSuchElementException 异常 |
| offer | 添加一个元素并返回true | 如果队列已满，则返回false |
| poll | 移除并返问队列头部的元素 | 如果队列为空，则返回null |
| peek | 返回队列头部的元素 | 如果队列为空，则返回null |
| put | 添加一个元素 | 如果队列满，则阻塞 |
| take | 移除并返回队列头部的元素 | 如果队列为空，则阻塞 |
class Solution {
public int[][] matrixReshape(int[][] nums, int r, int c) {
int row = nums.length;
int col = nums[0].length;
if(row*col!=r*c){return nums;}
Queue queue = new LinkedList<>();//创建一个队列
for(int i=0;i 方法三：方法三：用两个变量来记录当前所在行列，不需要额外的二维数组空间，直接将其放入
class Solution {
public int[][] matrixReshape(int[][] nums, int r, int c) {
int row = nums.length;
int col = nums[0].length;
if(row==0||row*col!=r*c){return nums;}
int[][] arr = new int[r][c];
int curRow=0,curCol=0;//定义两个变量，记录当前行与列
for(int i=0;i方法四：定义一个比变量k，用k去记录当前所在位置，k/c和k%c去找到当前行与列
class Solution {
public int[][] matrixReshape(int[][] nums, int r, int c) {
if(nums.length*nums[0].length!=r*c||nums.length==0){return nums;}
int[][] arr = new int[r][c];
int k=0;
for(int i=0;i