[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

---

给你一个 2 行 n 列的二进制数组：

• 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。

你需要利用 upper，lower 和 colsum 来重构这个矩阵，并以二维整数数组的形式返回它。

如果有多个不同的答案，那么任意一个都可以通过本题。

如果不存在符合要求的答案，就请返回一个空的二维数组。

示例 1：

输入：upper = 2, lower = 1, colsum = [1,1,1]
输出：[[1,1,0],[0,0,1]]
解释：[[1,0,1],[0,1,0]] 和 [[0,1,1],[1,0,0]] 也是正确答案。

示例 2：

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

示例 3：

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

提示：

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

---

Runtime: 904 ms

Memory Usage: 31.5 MB

class Solution {
    func reconstructMatrix(_ upper: Int, _ lower: Int, _ colsum: [Int]) -> [[Int]] {
        var upper = upper
        let lower = lower
        if upper + lower != colsum.reduce(0,+)
        {
            return [[Int]]()
        }
        var a:[Int] = [Int]()
        var b:[Int] = [Int]()
        for i in colsum
        {
            if upper > 0 && i != 0
            {
                upper -= 1
                a += [1]
            }
            else
            {
                a += [0]   
            }
            b += [i - a.last!]
        }
        if upper == 0
        {
            return [a,b]
        }
        return [[Int]]()
    }
}

---

Runtime: 728 ms

Memory Usage: 29.3 MB

class Solution {
    func reconstructMatrix(_ upper: Int, _ lower: Int, _ colsum: [Int]) -> [[Int]] {

        var indices = [Int]()
        var onesCount = 0
        var  ans = Array<Array<Int>>(repeating: Array<Int>(repeating: 0, count: colsum.count), count: 2)
        for i in 0..<colsum.count {
            if colsum[i] == 2 {
                ans[1][i] = 1
                ans[0][i] = 1
                onesCount += 1
            } else if colsum[i] == 1 {
                indices.append(i)
            }
        }
        let upperRemain = upper - onesCount
        let lowerRemain = lower - onesCount
        guard upperRemain >= 0 && lowerRemain >= 0 && upperRemain + lowerRemain == indices.count else {
            return []
        }
        let currentUpperSum = upper - onesCount
        var j = 0
        while j < currentUpperSum {
            ans[0][indices[j]] = 1
            j += 1
        }
        while j < indices.count {
            ans[1][indices[j]] = 1
            j += 1
        }
        return ans
    }
}