Li Kou daily question - day 28 -566 Reshape matrix

2022.6.26 Did you brush the questions today ？

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subject ：

stay MATLAB in , There's a very useful function reshape , It can put a  m x n The matrix is remodeled to another different size （r x c） The new matrix of , But keep its original data .

Give you a two-dimensional array mat It means  m x n matrix , And two positive integers r and c , Represent the number of rows and columns of the matrix to be reconstructed respectively .

The reconstructed matrix needs to replace all elements of the original matrix with the same Row traversal order fill .

If the reshape The operation is feasible and reasonable , The new reshaping matrix is output ; otherwise , Output raw matrix .

analysis ：

The meaning of this question is to give you a two-dimensional array , Then I'll give you another line , Column size , You need to judge the new line , Whether the column size can form a new array under the size of the original two-dimensional array , Return a new array if possible , Otherwise, the original two-dimensional array is returned , for example .

【【1,2】,【3,4】】,r=1,c=4, So you can construct 【1,2,3,4】

【【1,2】,【3,4】】,r=1,c=3 Then you can't form a new array

Ideas . For reshaping matrix types , We all need to treat the original array as a one-dimensional array , And then use it ,/ and % Two residual methods are used for assignment .

analysis ：

1. Violent solution

class Solution {
public:
    vector<vector<int>> matrixReshape(vector<vector<int>>& mat, int r, int c) {
    vector<vector<int>>vec(r, vector<int>(c));
    if (r * c != mat.size()*mat[0].size())
    {
        return mat;
    }
    else
    {
        for (auto i = 0; i < mat.size() * mat[0].size(); i++)
        {
            vec[i / c][i % c] = mat[i / mat[0].size()][i % mat[0].size()];
        }
    }
    return vec;
    }
};