LeetCode笔记：566. Reshape the Matrix

问题（Easy）：

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.

大意：

在MATLAB中，有一个很有用的函数名为“reshape”，可以重构一个矩阵为另一个尺寸，并保持原始数据。

给你一个由二维数组表示的矩阵，和两个正数r和c，分别表示想要重构成的新矩阵的行数和列数。

重构的矩阵需要由所有原来矩阵的元素以同样的顺序填充。

如果根据给出的参数进行重构操作是可能和合法的，就输出重构出的新矩阵，否则就输出原始矩阵。

例1：
输入：
nums =
[[1,2],
[3,4]]
r = 1, c = 4
输出：
[[1,2,3,4]]
解释：
原矩阵的顺序是[1,2,3,4]。新重构的是个1*4的矩阵，可以用上面的列表来一行行填充。

例2：
输入：
nums =
[[1,2],
[3,4]]
r = 2, c = 4
输出：
[[1,2],
[3,4]]
解释：
无法将22的矩阵重构为24的矩阵。因此输出原始矩阵。

注意：

1. 给出的矩阵高宽在[1,100]范围内。
2. 给出的r和c是正数。

思路：

也没什么特别的思路，就是遍历原二维数组，来按照数量建立新的二位数组，C++中用容器实现。唯一要注意的就是操作前的参数判断：是否为空数组、是否元素数一致、是否没变化之类的。

代码（C++）：

class Solution {
public:
    vector> matrixReshape(vector>& nums, int r, int c) {
        if (nums.size() == 0) return nums;
        if ((r * c) != (nums.size() * nums[0].size())) return nums;
        if (r == nums.size()) return nums;
        
        vector> res;
        int oldC = 0;
        int oldR = 0;
        for (int i = 0; i < r; i++) {
            vector newRow;
            for (int j = 0; j < c; j++) {
                if (oldC < nums[0].size()) {
                    newRow.push_back(nums[oldR][oldC]);
                    oldC++;
                } else {
                    oldR++;
                    oldC = 0;
                    newRow.push_back(nums[oldR][oldC]);
                    oldC++;
                }
            }
            res.push_back(newRow);
        }
        return res;
    }
};

他山之石：

别人的代码就简洁很多：

class Solution {
public:
    vector> matrixReshape(vector>& nums, int r, int c) {
        int m = nums.size(), n = nums[0].size(), o = m * n;
        if (r * c != o) return nums;
        vector> res(r, vector(c, 0));
        for (int i = 0; i < o; i++) res[i / c][i % c] = nums[i / n][i % n];
        return res;
    }
};

合集：https://github.com/Cloudox/LeetCode-Record

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