leetcode566. Reshape the Matrix

最新推荐文章于 2021-05-22 20:29:00 发布

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最新推荐文章于 2021-05-22 20:29:00 发布

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本文介绍了一种将矩阵重塑为不同尺寸但保持原始数据的算法。通过两个示例详细展示了重塑过程，并提供了两种不同的实现方法，包括一种优化版本。

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566. Reshape the Matrix

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.

解法

第几个元素 / 列数是所在的行数，% 列数是所在的列数。

public class Solution {
    public int[][] matrixReshape(int[][] nums, int r, int c) {
        if (nums == null || nums.length == 0) {
            return null;
        }
        if (nums[0] == null || nums[0].length == 0) {
            return null;
        }

        int x = nums.length;
        int y = nums[0].length;
        int len = x * y;
        if (len != r * c) {
            return nums;
        }
        int[][] reshape = new int[r][c];
        for (int i = 0; i < r; i++) {
            for (int j = 0; j < c; j++) {
                int temp = i * c + j;
                int position_x = temp / y;
                int position_y = temp % y;
                reshape[i][j] = nums[position_x][position_y];
            }
        }

        return reshape;
    }
}

解法二

解法一的优化版本

public class Solution {
    public int[][] matrixReshape(int[][] nums, int r, int c) {
        if (nums == null || nums.length == 0 || r * c != nums.length * nums[0].length) {
            return nums;
        }
        if (nums[0] == null || nums[0].length == 0) {
            return nums;
        }

        int[][] reshape = new int[r][c];
        int count = 0;
        for (int i = 0; i < r; i++) {
            for (int j = 0; j < c; j++) {
                reshape[i][j] = nums[count / nums[0].length][count % nums[0].length];
                count++;
            }
        }

        return reshape;
    }
}