LeetCode 566. Reshape the Matrix (C++)

题目：

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.

分析：

给定一个矩阵，根据参数返回一个新的矩阵。

可以发现，只有给的r * c和原来矩阵行列乘积相同才可以将矩阵正确的变形，否则返回原来的矩阵。用f来确定原矩阵中元素的索引。直接两重循环将元素填进新矩阵中。

程序：

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