leetcode 566. Reshape the Matrix 重塑矩阵

最新推荐文章于 2021-05-22 20:29:00 发布

JackZhangNJU

最新推荐文章于 2021-05-22 20:29:00 发布

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分类专栏： leetcode For C++

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本文介绍了一个简单的矩阵操作算法——重塑矩阵。该算法接受一个二维数组作为输入，并将它转换为指定行列数的新矩阵，同时保持元素原有的行遍历顺序。若无法完成重塑，则返回原始矩阵。文中提供了一个C++实现示例。

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.

本题题意很简单，直接遍历矩阵即可

代码如下：

#include <iostream>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <string>
#include <climits>
#include <algorithm>
#include <sstream>
#include <functional>
#include <bitset>
#include <numeric>
#include <cmath>

using namespace std;


class Solution 
{
public:
    vector<vector<int>> matrixReshape(vector<vector<int>>& nums, int r, int c) 
    {
        int row = nums.size(), col = nums[0].size();
        if (row*col != r*c)
            return nums;

        vector<vector<int>> res(r, vector<int>(c, 0));
        for (int i = 0; i < r; i++)
        {
            for (int j = 0; j < c; j++)
            {
                int k = i*c + j;
                res[i][j] = nums[k/col][k%col];
            }
        }
        return res;
    }
};