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<row;++i){
            for(int j=0;j<col;++j){
                arr1[k++]=nums[i][j];//一位数组将所有的数依次存入
            }
        }
        k=0;
        int[][] arr = new int[r][c];
        for(int i=0;i<r;++i){
            for(int j=0;j<c;++j){
                arr[i][j]=arr1[k++];//根据所谓的长和宽依次reshape，每隔一个c的长度一个内循环
            }
        }
        return arr;
    }
}

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

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<Integer> queue = new LinkedList<>();//创建一个队列
        for(int i=0;i<row;++i){
            for(int j=0;j<col;++j){
                queue.add(nums[i][j]);//添加元素
            }
        }
        int[][] arr = new int[r][c];
        for(int i=0;i<r;++i){
            for(int j=0;j<c;++j){
                arr[i][j]=queue.remove();//从队列头获取并删除一个元素
            }
        }
        return arr;
    }
}

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<nums.length;++i){
            for(int j=0;j<nums[0].length;++j){
                arr[curRow][curCol]=nums[i][j];//进行赋值
                curCol++;
                if(curCol==c){//当到达每一行的结尾
                    curRow++;
                    curCol=0;
                }
            }
        }
        return arr;
    }
}

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<nums.length;++i){
            for(int j=0;j<nums[0].length;++j){
                arr[k/c][k%c]=nums[i][j];//将数组中值通过计算，放到新数组中
                k++;
            }
        }
        return arr;
    }
}