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.

问题描述：就是矩阵里经常有的reshape操作，把一个数组改变行与列的个数，变成另外一个数组。 解题思路：首先看看转换前的数组与转换后的数组的元素个数是否相匹配，如果不匹配则返回原数组；如果匹配，则一个一个传值过去。 代码如下

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;
    }
};