566. Reshape the Matrix

Posted 2020-10-10 andywu

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:

1. The height and width of the given matrix is in range [1, 100].
2. The given r and c are all positive.

题意： 将原矩阵变换成r行c列

思路：将矩阵变为一维数组，然后再变换为r*c矩阵

 1     public int[][] matrixReshape(int[][] nums, int r, int c) {
 2         int rows = nums.length;
 3         if (rows == 0 || r == 0 || c == 0) {
 4             return nums;
 5         }
 6         int cols = nums[0].length;
 7 
 8         if (rows * cols != r * c) {
 9             return nums;
10         }
11 
12         int []values = new int[rows*cols];
13         int index=0;
14         for (int row = 0; row < rows; row++)
15             for (int col = 0; col < cols; col++)
16             {
17                 values[index++] = nums[row][col];
18             }
19 
20         int[][] newNums = new int[r][c];
21         index=0;
22         for (int i = 0; i < r; i++)
23             for (int j = 0; j < c; j++) {
24                 newNums[i][j] = values[index++];
25             }
26         return newNums;
27     }