[LeetCode] Reshape the Matrix 重塑矩阵

2023年12月11日 244次阅读 来源: Grandyang

 

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.

 

这道题让我们实现矩阵大小的重塑，也就是实现Matlab中的reshape函数，博主也经常使用matlab，对这个函数还是比较的熟悉的。对于这种二维数组大小重新非配的问题的关键就是对应位置的坐标转换，最直接的办法就是先把原数组拉直，变成一条直线，然后再组成新的数组。所以这道题我们先判断给定数组是否能重塑成给定的大小，就是看两者的元素总数是否相同，直接行数乘以列数即可，然后我们新建一个目标大小的数组，并开始遍历，对于每个位置，我们先转为拉直后的一维坐标，然后在算出在原数组中的对应位置赋值过来即可，参见代码如下：

 

解法一：

class Solution {
public:
    vector<vector<int>> matrixReshape(vector<vector<int>>& nums, int r, int c) {
        int m = nums.size(), n = nums[0].size();
        if (m * n != r * c) return nums;
        vector<vector<int>> res(r, vector<int>(c));
        for (int i = 0; i < r; ++i) {
            for (int j = 0; j < c; ++j) {
                int k = i * c + j;
                res[i][j] = nums[k / n][k % n];
            }
        }
        return res;
    }
};

 

下面这种方法整体思路和上面没啥区别，但是只使用了一个循环，直接就是遍历拉直后的一维数组的坐标，然后分别转换为两个二维数组的坐标进行赋值，参见代码如下：

 

解法二：

class Solution {
public:
    vector<vector<int>> matrixReshape(vector<vector<int>>& nums, int r, int c) {
        int m = nums.size(), n = nums[0].size();
        if (m * n != r * c) return nums;
        vector<vector<int>> res(r, vector<int>(c));
        for (int i = 0; i < r * c; ++i) {
            res[i / c][i % c] = nums[i / n][i % n];
        }
        return res;
    }
};

 

参考资料：

https://discuss.leetcode.com/topic/87851/java-concise-o-nm-time

 

    原文作者：Grandyang
    原文地址: http://www.cnblogs.com/grandyang/p/6804753.html
    本文转自网络文章，转载此文章仅为分享知识，如有侵权，请联系博主进行删除。
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