LeetCode 566. Reshape the Matrix (Easy)

矩阵重塑算法

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本文介绍了一个在MATLAB中常用的矩阵重塑算法，并提供了C++实现代码。该算法能够将一个一维或二维数组转换为指定行数和列数的新矩阵，同时保持原有数据不变。若无法按指定大小重塑，则返回原数组。

题目描述：

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.

Example1：

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.

Example2：

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.

题目大意：给一个数组，将它重塑成二维数组，如果输入的重塑大小与原数组大小不符（元素总数不同），那么输出原数组。

思路：先判断能不能重塑，再直接模拟一下即可。

c++代码：

class Solution {
public:
    vector<vector<int>> matrixReshape(vector<vector<int>>& nums, int r, int c) {
        if (nums.size() > 0 && nums[0].size() > 0 && nums.size() * nums[0].size() == r * c)
        {
            vector<vector<int>> ret;
            int count = 0;
            for (size_t i = 0; i < r; ++i)
            {
                vector<int> temp;
                for (size_t j = 0; j < c; ++j)
                {
                    temp.push_back(nums[count / nums[0].size()][count % nums[0].size()]);
                    ++count;
                }
                ret.push_back(temp);
            }
            return ret;
        }
        return nums;
    }
};