leetcode week13

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.

問題描述：就是矩陣裡經常有的reshape操作，把一個數組改變行與列的個數，變成另外一個數組。 解題思路：首先看看轉換前的數組與轉換後的數組的元素個數是否相比對，如果不比對則傳回原數組；如果比對，則一個一個傳值過去。 代碼如下

class Solution {
public:
    vector<vector<int>> matrixReshape(vector<vector<int>>& nums, int r, int c) {
        int old_c = nums[0].size();
        int old_r = nums.size();
        
        if (old_c * old_r != r * c)
            return nums;

        vector<vector<int>> reshaped(r, vector<int>(c));
        int old_i = 0; // row
        int old_j = 0; // col
        for (int i = 0; i < r; ++i) // row
        {
            for (int j = 0; j < c; ++j) // col
            {
                reshaped[i][j] = nums[old_i][old_j++];
                if (old_j == old_c)
                {
                    old_j = 0;
                    ++old_i;
                }
            }
        }
        return reshaped;
    }
};