（leetcode题解）Reshape the Matrix

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.

题意将一个矩阵的行列改变，但不改变里面的数值。

我的做法是直接遍历原矩阵赋给新矩阵，C++实现如下

vector<vector<int>> matrixReshape(vector<vector<int>>& nums, int r, int c) {
        int row=nums.size();
        int vol=nums[0].size();
        vector<vector<int>> res(r,vector<int>(c,0));
        int temp=row*vol;
        if(temp!=r*c)
            return nums;
        int m=0,n=0;
        for(int i=0;i<row;i++)
        {
            for(int j=0;j<vol;j++)
            {
                res[m][n++]=nums[i][j];
                if(n>=c)
                {
                    m++;
                    n=0;
                }
            }
        }
        return res;
    }

我看了一下网上有更简洁的做法，不过时间复杂度没改变还是O（r*c）,贴下来以供参考

    vector<vector<int>> matrixReshape(vector<vector<int>>& nums, int r, int c) {
        int row=nums.size();
        int vol=nums[0].size();
        vector<vector<int>> res(r,vector<int>(c,0));
        int temp=row*vol;
        if(temp!=r*c)
            return nums;
        for(int i=0;i<temp;i++)
        {
            res[i/c][i%c]=nums[i/vol][i%vol];
        }
        return res;
    }

（leetcode题解）Reshape the Matrix

原文：http://www.cnblogs.com/kiplove/p/6964508.html