## [Swift]LeetCode1253. 重构 2 行二进制矩阵 | Reconstruct a 2-Row Binary Matrix

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Given the following details of a matrix with `n` columns and `2` rows :

• The matrix is a binary matrix, which means each element in the matrix can be `0` or `1`.
• The sum of elements of the 0-th(upper) row is given as `upper`.
• The sum of elements of the 1-st(lower) row is given as `lower`.
• The sum of elements in the i-th column(0-indexed) is `colsum[i]`, where `colsum` is given as an integer array with length `n`.

Your task is to reconstruct the matrix with `upper``lower` and `colsum`.

Return it as a 2-D integer array.

If there are more than one valid solution, any of them will be accepted.

If no valid solution exists, return an empty 2-D array.

Example 1:

```Input: upper = 2, lower = 1, colsum = [1,1,1]
Output: [[1,1,0],[0,0,1]]
Explanation: [[1,0,1],[0,1,0]], and [[0,1,1],[1,0,0]] are also correct answers.
```

Example 2:

```Input: upper = 2, lower = 3, colsum = [2,2,1,1]
Output: []
```

Example 3:

```Input: upper = 5, lower = 5, colsum = [2,1,2,0,1,0,1,2,0,1]
Output: [[1,1,1,0,1,0,0,1,0,0],[1,0,1,0,0,0,1,1,0,1]]```

Constraints:

• `1 <= colsum.length <= 10^5`
• `0 <= upper, lower <= colsum.length`
• `0 <= colsum[i] <= 2`

• 矩阵是一个二进制矩阵，这意味着矩阵中的每个元素不是 `0` 就是 `1`
• 第 `0` 行的元素之和为 `upper`
• 第 `1` 行的元素之和为 `lower`
• 第 `i` 列（从 `0` 开始编号）的元素之和为 `colsum[i]``colsum` 是一个长度为 `n` 的整数数组。

```输入：upper = 2, lower = 1, colsum = [1,1,1]

```

```输入：upper = 2, lower = 3, colsum = [2,2,1,1]

```

```输入：upper = 5, lower = 5, colsum = [2,1,2,0,1,0,1,2,0,1]

• `1 <= colsum.length <= 10^5`
• `0 <= upper, lower <= colsum.length`
• `0 <= colsum[i] <= 2`