Reshape the Matrix

• LeetCode: Reshape the Matrix
• Problem:

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.

• Note:

1. The height and width of the given matrix is in range [1, 100].

2. The given r and c are all positive.

• Exmaples:

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.

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.

• Analysis:

The size of the reshaped matrix is equal to the size of the original matrix can be reshaped.

• Code - Java:

public class Solution {
    public int[][] matrixReshape(int[][] nums, int r, int c) {
        int row = nums.length;
        if (row == 0) {
            return nums;
        }
        int col = nums[0].length;
        if (col == 0) {
            return nums;
        }
        int newSize = r * c;
        int currentSize = row * col;
        if (newSize != currentSize) {
            return nums;
        }
        int[][] result = new int[r][c];
        int k = 0;
        int l = 0;
        for (int i = 0; i < r; i++) {
            for (int j = 0; j < c; j++) {
                result[i][j] = nums[k][l];
                if (l == col - 1) {
                    k++;
                }
                l = (l + 1) % col;
            }
        }
        return result;
    }
}

• Code - C++:

class Solution {
public:
    vector<vector<int>> matrixReshape(vector<vector<int>>& nums, int r, int c) {
        int row = nums.size();
        if (row == 0) {
            return nums;
        }
        int col = nums[0].size();
        if (col == 0) {
            return nums;
        }
        int newSize = r * c;
        int currentSize = row * col;
        if (newSize != currentSize) {
            return nums;
        }
        vector<vector<int>> result(r, vector<int>(c, 0));
        int k = 0;
        int l = 0;
        for (int i = 0; i < r; i++) {
            for (int j = 0; j < c; j++) {
                result[i][j] = nums[k][l];
                if (l == col - 1) {
                    k++;
                }
                l = (l + 1) % col;
            }
        }
        return result;
    }
};

• Code - Python:

class Solution(object):
    def matrixReshape(self, nums, r, c):
        """
        :type nums: List[List[int]]
        :type r: int
        :type c: int
        :rtype: List[List[int]]
        """
        row = len(nums)
        if row == 0:
            return nums
        col = len(nums[0])
        if col == 0:
            return nums
        newSize = r * c
        currentSize = row * col
        if newSize != currentSize:
            return nums
        result = [[0 for j in range(c)] for i in range(r)]
        k, l = 0, 0
        for i in range(r):
            for j in range(c):
                result[i][j] = nums[k][l]
                if l == col - 1:
                    k += 1
                l = (l + 1) % col
        return result

• Time Complexity: O(MN), M and N are the row and column of the original matrix.
• Space Complexity: O(MN), M and N are the row and column of the original matrix.