LeetCode566. Reshape the Matrix

Description

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.

my program

思路：先判断能否reshape，判断标准是原矩阵的行×列是否等于reshape后的矩阵的行×列。
然后一层循环一个一个复制赋值即可。关键是要控制好矩阵的行列下标。
时间复杂度O(rc)

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

Submission Details
56 / 56 test cases passed.
Status: Accepted
Runtime: 42 ms

//双循环，时间复杂度O(rc)
class Solution {
public:
    vector<vector<int>> matrixReshape(vector<vector<int>>& nums, int r, int c) {
        int m = nums.size(), n = nums[0].size();
        if (m * n != r * c) {
            return nums;
        }

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

        return res;
    }
};