[Swift]LeetCode1128. 等价多米诺骨牌对的数量 | Number of Equivalent Domino Pairs

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Given a list of dominoes, dominoes[i] = [a, b] is equivalent to dominoes[j] = [c, d] if and only if either (a==c and b==d), or (a==d and b==c) - that is, one domino can be rotated to be equal to another domino.

Return the number of pairs (i, j) for which 0 <= i < j < dominoes.length, and dominoes[i] is equivalent to dominoes[j].

Example 1:

Input: dominoes = [[1,2],[2,1],[3,4],[5,6]]
Output: 1

Constraints:

• 1 <= dominoes.length <= 40000
• 1 <= dominoes[i][j] <= 9

---

给你一个由一些多米诺骨牌组成的列表 dominoes。

如果其中某一张多米诺骨牌可以通过旋转 0 度或 180 度得到另一张多米诺骨牌，我们就认为这两张牌是等价的。

形式上，dominoes[i] = [a, b] 和 dominoes[j] = [c, d] 等价的前提是 a==c 且 b==d，或是 a==d 且 b==c。

在 0 <= i < j < dominoes.length 的前提下，找出满足 dominoes[i] 和 dominoes[j] 等价的骨牌对 (i, j) 的数量。

示例：

输入：dominoes = [[1,2],[2,1],[3,4],[5,6]]
输出：1

提示：

• 1 <= dominoes.length <= 40000
• 1 <= dominoes[i][j] <= 9

---

192ms

class Solution {
    func dominoHash(_ dom:[Int]) -> Int {
        if dom[0] > dom[1] {
            return 100*dom[1] + dom[0]
        }
        return 100*dom[0] + dom[1]
    }

    func numEquivDominoPairs(_ dominoes: [[Int]]) -> Int {
        var result = 0
        var counts = [Int : Int]()
        
        for d in dominoes {
            let h = dominoHash(d)
            if let count = counts[h] {
                result += count
                counts[h] = count + 1
            } else {
                counts[h] = 1
            }
        }
        return result
    }
}

---

196ms

sample 196 ms submission
class Solution {
    func numEquivDominoPairs(_ dominoes: [[Int]]) -> Int {
        var originals = [Int]()

        dominoes.forEach { (domino) in
            originals.append(10*min(domino[0], domino[1]) + max(domino[0], domino[1]))
        }

        var counts: [Int: Int] = [:]
        originals.forEach { counts[$0, default: 0] += 1 }

        var count = 0
        counts.forEach { count = count + ( $1 * ($1 - 1)/2 ) }

        return count
    }
}

---

200ms

class Solution {
    func numEquivDominoPairs(_ dominoes: [[Int]]) -> Int {
        let dominoes = dominoes.map { dom -> Int in
            if dom[0] <= dom[1] {
                return dom[0] * 10 + dom[1]
            } else {
                return dom[1] * 10 + dom[0]
            }
        }
        var counts = [Int: Int]()
        var res = 0
        for num in dominoes {
            res += (counts[num] ?? 0)
            counts[num, default: 0] += 1
        }
        return res
    }
}

---

Runtime: 224 ms

Memory Usage: 23.9 MB

class Solution {
    func numEquivDominoPairs(_ dominoes: [[Int]]) -> Int {
        var dominoes = dominoes
        var freq:[[Int]:Int] = [[Int]:Int]()
        var total:Int = 0
        for i in 0..<dominoes.count
        {
            if dominoes[i][0] > dominoes[i][1]
            {
                dominoes[i].swapAt(0,1)
            }
            total += freq[[dominoes[i][0], dominoes[i][1]],default:0]
            freq[[dominoes[i][0], dominoes[i][1]],default:0] += 1
        }
        return total
    }
}