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.

遍历原二维数组依次取出数据放到新数组中

看到这道题的时候我想到了这个[LeetCode] 806. Number of Lines To Write String，于是我用了同样的思路，维护一个col_tmp记录当前的列下标，到了新数组的列数就换行

vector<vector<int>> matrixReshape(vector<vector<int>>& nums, int r, int c)
{
    int row = nums.size();
    int col = nums.at(0).size();

    if (row * col != r * c)
    {
        return nums;
    }

    vector<vector<int>> result;
    vector<int> tmp;

    int col_tmp = 0;

    for (int i = 0; i < row; i++)
    {
        for (int j = 0; j < col; j++)
        {
            if (tmp.size() + 1 < c)
            {
                tmp.push_back(nums.at(i).at(j));
                ++col_tmp;
            }
            else
            {
                tmp.push_back(nums.at(i).at(j));
                result.push_back(tmp);
                tmp.clear();
                col_tmp = 0;
            }
        }
    }

    return result;
}

看看LeetCode上其他代码，利用/和%算出reshape后的下标，简洁明了

vector<vector<int>> matrixReshape(vector<vector<int>>& nums, int r, int c) {
    if (nums.empty() || nums[0].empty()) {
        return nums;
    }

    size_t rows = nums.size();
    size_t columns = nums[0].size();
    size_t input_size = rows * columns;
    size_t traversing_size = r * c;

    if (traversing_size > input_size) {
        // illegal
        return nums;
    }

    vector<vector<int>> traversing(r, vector<int>(c, 0));

    //填满新的二维数组
    for (int k = 0; k < r; k++) {
        for (int m = 0; m < c; m++) {
            // 算出在原数组中的下标
            int linear_address = k * c + m;
            int i = linear_address / columns;
            int j = linear_address % columns;
            traversing[k][m] = nums[i][j];
        }
    }

    return traversing;
}