566. Reshape the Matrix

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本文介绍了一种将矩阵从原始尺寸重塑为目标尺寸的算法实现。通过行优先遍历原矩阵元素并将其转换为一维数组，再根据目标尺寸重塑为新的二维矩阵。文章提供了具体的Java代码示例。

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.

題意:

將一個矩陣從m*n維變形為r*c維，其中m*n=r*c，若發生下面情況則返回原矩陣:

m=r且n=c

m*n != r*c

例如:

2*3 3*2

============

1 2 3 1 2

4 5 6 => 3 4

5 6

題解:

先將m*n的2維矩陣變成1維作為中間矩陣，然後將這個1維的矩陣轉換為r*c的2維矩陣，再返回這個r*c的2維矩陣即可。

package LeetCode.Easy;

public class ReshapeTheMatrix {
    public int[][] matrixReshape(int[][] nums, int r, int c) {
        //排除其他情況
        if(nums == null || nums.length == 0 || nums[0].length == 0)
            return nums;
        
        if(r <= 0 || c <= 0)
            return nums;
        
        int m = nums.length;
        int n = nums[0].length;
        
        //兩個矩陣形狀無法進行轉換
        if(r * c != m * n)
            return nums;
        
        //若兩個矩陣長寬都一樣就不用變形了
        if(r == m && c == n)
            return nums;
        
        //正式開始進行矩陣的reshape
        
        //先將矩陣變成一維
        int[] temp = new int[m * n];
        int cnt = 0;
        for(int i = 0; i < m; i ++) {
            for(int j = 0; j < n; j ++) {
                temp[cnt] = nums[i][j];
                cnt ++;
            }
        }
        
        //再將一維矩陣轉換為r*c的矩陣
        int[][] ans = new int[r][c];
        cnt = 0;
        for(int i = 0; i < r; i ++) {
            for(int j = 0; j < c; j ++) {
                ans[i][j] = temp[cnt];
                cnt ++;
            }
        }
        
        return ans;
    }
}