566. Reshape the Matrix

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本文介绍了一种将矩阵重新塑形的方法，确保元素数量一致的前提下改变其行数和列数。通过遍历原始矩阵并按行顺序填充到新矩阵中实现转换。

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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

要按照输入将一个矩阵转化为另一个矩阵，前提是这两个矩阵的元素个数相同

算法思路

首先拿到原来数组的行列个数，与输入的判断，元素个数不相同直接返回原数组

然后对所有的数据进行遍历，新数组的行列分别对应i除以列数和取余列数，同理原数组的行列关系

代码：

package leetcode;

public class ReshapeTheMatrix {
	
	public int[][] matrixReshape(int[][] nums, int r, int c) {		
		int org_r = nums.length;
		int org_c = nums[0].length;
		int[][] result = new int[r][c];
		if(org_r * org_c != r * c){
			
			return nums;
		}else{
			for(int i=0; i<r*c; i++){
				result[i/c][i%c] = nums[i/org_c][i%org_c];
			}
			
			return result;
		}
        
    }
	
	
}