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.

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.

思路

这个题就是把二维数组“捋直”，变成一维数组，然后按新的行列进行分配。不过不用真正进行“捋直”，只要知道他们是怎么转换的即可。

代码一（思路不够简洁）

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

代码二（思路简洁，和一思想一致）

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