566. Reshape the Matrix

QUESTION:

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 columnnumber 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.

EXPLANATION

其实就是计算面积的，然后再重新填充。

刚开始的时候以为不用保存原始的数据，所以直接使用了count++的方式，后来才发现使用原始数据。当时也没有想到有啥更好的办法，于是一股脑儿将数据都输入进去然后再输出出来。

然后看了下别人的submit，发现原来数据总数是一定的，那么就可以遍历数据挨个填充即可。

确实优雅和只需要一个循环。

SOLUTION:

public class Solution {
    public int[][] matrixReshape(int[][] nums, int r, int c) {
        if (nums.length * nums[0].length != r * c) return nums;
        int[][] result = new int[r][c];
        Queue input = new LinkedList<Integer>();
        for(int i = 0;i<nums.length;i++){
            for(int j = 0;j<nums[i].length;j++){
                input.add(nums[i][j]);
            }
        }
        for(int i = 0;i<r;i++){
            for(int j =0;j<c;j++){
                result[i][j] = (int) input.remove();
            }
        }
        return result;
    }
}
//我看到比较好的解决方法
public class Solution {
    public int[][] matrixReshape(int[][] nums, int r, int c) {
        int m = nums.length, n = m > 0 ? nums[0].length : 0;
        if (r * c != m * n)
            return nums;
        int[][] reshaped = new int[r][c];
        for (int i=0; i<m*n; i++)
            reshaped[i/c][i%c] = nums[i/n][i%n];
        return reshaped;
    }
}
--> return <--