19.leetCode566:Reshape the Matrix

题目：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.

题意：给定一个m*n的矩阵，重塑该矩阵，使之变成x*y（m*n=x*y），但保持其中的元素及元素顺序不变。

思路一：使用一个queue，遍历旧的矩阵，将元素都放入queue中。再遍历一遍新的矩阵，把queue中的元素按新矩阵的次序放好。

思路二：遍历一次旧矩阵，就把值直接赋值给新矩阵中对应的位置。
代码：

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[][] result = new int[r][c];        for(int i=0;i<row;i++){            for(int j=0;j<column;j++){                int num = i*column+ j;                int m = num/c;                int n = num%c;                result[m][n] = nums[i][j];            }        }        return result;    }}