Leetcode 566(Java)

最新推荐文章于 2022-10-19 20:14:18 发布

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最新推荐文章于 2022-10-19 20:14:18 发布

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本文介绍了一个基于给定矩阵重塑为不同尺寸新矩阵的算法。该算法遵循特定的填充顺序，并在无法重塑时返回原始矩阵。文章通过两个示例详细解释了这一过程。

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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.

首先判断要输出结果的矩阵大小与原矩阵大小是否相符，若不相符返回原矩阵即可。判定方法也很简单，只要元素个数相等就可以视为符合。接下来利用队列，将原矩阵的元素一个个ADD进去，再一个个POLL到新矩阵中，输出即可。

AC码：

public class Solution {
    public int[][] matrixReshape(int[][] nums, int r, int c) {
        int rr = nums.length,cc=nums[0].length;
        int[][] result = new int[r][c];
        if(rr*cc != r*c)return nums;
        Queue<Integer> queue = new LinkedList<Integer>();
        for(int i=0;i<rr;i++){
            for(int j=0;j<cc;j++){
                queue.add(nums[i][j]);
            }
        }
        for(int x=0;x<r;x++){
            for(int y=0;y<c;y++){
                result[x][y] = queue.poll();
            }
        }
        return result;
    }
}