566. Reshape the Matrix

566. Reshape the Matrix (Easy)

In MATLAB, there is a handy function called reshape which can reshape an m x n matrix into a new one with a different size r x c keeping its original data.

You are given an m x n matrix mat and two integers r and c representing the row number and column number of the wanted reshaped matrix.

The reshaped matrix should 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: mat = [[1,2],[3,4]], r = 1, c = 4
Output: [[1,2,3,4]]

Example 2:

Input: mat = [[1,2],[3,4]], r = 2, c = 4
Output: [[1,2],[3,4]]

 

Constraints:

• m == mat.length
• n == mat[i].length
• 1 <= m, n <= 100
• -1000 <= mat[i][j] <= 1000
• 1 <= r, c <= 300

Companies:
Mathworks

Related Topics:
Array, Matrix, Simulation

Solution 1.

// OJ: https://leetcode.com/problems/reshape-the-matrix/
// Author: github.com/lzl124631x
// Time: O(MN)
// Space: O(1)
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>> ans;
        for (int i = 0; i < r; ++i) {
            vector<int> row;
            for (int j = 0; j < c; ++j) {
                int index = i * c + j;
                row.push_back(nums[index / N][index % N]);
            }
            ans.push_back(row);
        }
        return ans;
    }
};

Or

// OJ: https://leetcode.com/problems/reshape-the-matrix/
// Author: github.com/lzl124631x
// Time: O(MN)
// Space: O(1)
class Solution {
public:
    vector<vector<int>> matrixReshape(vector<vector<int>>& A, int r, int c) {
        int M = A.size(), N = A[0].size(), a = 0, b = 0;
        if (M == r || r * c != M * N) return A;
        vector<vector<int>> ans(r, vector<int>(c));
        for (int i = 0; i < M; ++i) {
            for (int j = 0; j < N; ++j) {
                ans[a][b] = A[i][j];
                if (++b == c) {
                    b = 0;
                    ++a;
                }
            }
        }
        return ans;
    }
};