[Swift]LeetCode954. 二倍数对数组 | Array of Doubled Pairs
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➤微信公众号：山青咏芝（shanqingyongzhi）
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Given an array of integers A with even length, return true if and only if it is possible to reorder it such that A[2 * i + 1] = 2 * A[2 * i] for every 0 <= i < len(A) / 2.
Example 1:
Input: [3,1,3,6]
Output: false
Example 2:
Input: [2,1,2,6]
Output: false
Example 3:
Input: [4,-2,2,-4]
Output: true
Explanation: We can take two groups, [-2,-4] and [2,4] to form [-2,-4,2,4] or [2,4,-2,-4].
Example 4:
Input: [1,2,4,16,8,4]
Output: false
Note:
0 <= A.length <= 30000A.lengthis even-100000 <= A[i] <= 100000
给定一个长度为偶数的整数数组 A，只有对 A 进行重组后可以满足 “对于每个 0 <= i < len(A) / 2，都有 A[2 * i + 1] = 2 * A[2 * i]” 时，返回 true；否则，返回 false。
示例 1：
输入：[3,1,3,6] 输出：false
示例 2：
输入：[2,1,2,6] 输出：false
示例 3：
输入：[4,-2,2,-4] 输出：true 解释：我们可以用 [-2,-4] 和 [2,4] 这两组组成 [-2,-4,2,4] 或是 [2,4,-2,-4]
示例 4：
输入：[1,2,4,16,8,4] 输出：false
提示：
0 <= A.length <= 30000A.length为偶数-100000 <= A[i] <= 100000
688ms
1 class Solution { 2 func canReorderDoubled(_ A: [Int]) -> Bool { 3 var d = [Int: Int]() 4 for a in A { 5 d[a, default: 0] += 1 6 } 7 let ps = Array(d.keys).filter {$0 > 0}.sorted() 8 let ns = Array(d.keys).filter {$0 < 0}.sorted {$0 > $1} 9 for p in ps { 10 //print(d) 11 guard let u = d[p], u >= 0 else { 12 return false 13 } 14 d[2*p, default: 0] -= u 15 d[p] = 0 16 } 17 for n in ns { 18 //print(d, n) 19 guard let u = d[n], u >= 0 else { 20 return false 21 } 22 d[2*n, default: 0] -= u 23 d[n] = 0 24 } 25 guard d[0, default: 0] % 2 == 0 else { 26 return false 27 } 28 d[0] = 0 29 let c = Array(d.values).filter {$0 != 0}.count 30 guard c == 0 else { 31 return false 32 } 33 return true 34 } 35 }
788ms
1 class Solution { 2 func canReorderDoubled(_ A: [Int]) -> Bool { 3 var d = [Int: Int]() 4 for a in A { 5 d[abs(a), default: 0] += 1 6 } 7 let ps = Array(d.keys).sorted() 8 for p in ps { 9 guard d[p*2, default: 0] >= d[p]! else { 10 return false 11 } 12 d[2*p, default: 0] -= d[p]! 13 } 14 return true 15 } 16 }
1076ms
1 class Solution { 2 func canReorderDoubled(_ A: [Int]) -> Bool { 3 if A.count == 0 { 4 return true 5 } 6 if A.count % 2 != 0 { 7 return false 8 } 9 10 let A = A.sorted() 11 12 var memo = [Int: Int]() 13 for i in 0..<A.count { 14 let current = A[i] 15 16 if let value = memo[current * 2] { 17 if value - 1 == 0 { 18 memo.removeValue(forKey: current * 2) 19 } else { 20 memo[current * 2] = value - 1 21 } 22 } else if current % 2 == 0, let value = memo[current / 2] { 23 if value - 1 == 0 { 24 memo.removeValue(forKey: current / 2) 25 } else { 26 memo[current / 2] = value - 1 27 } 28 } else { 29 memo[current] = (memo[current] ?? 0) + 1 30 } 31 32 } 33 34 return memo.count == 0 35 } 36 }
1100ms
1 class Solution { 2 func canReorderDoubled(_ A: [Int]) -> Bool { 3 var n:Int = A.count 4 var a:[Int] = [Int](repeating:0,count:n) 5 for i in 0..<n 6 { 7 if A[i] < 0 8 { 9 a[i] = -A[i]*100000000 10 } 11 else 12 { 13 a[i] = A[i] 14 } 15 } 16 a = a.sorted(by: <) 17 var p:Int = 0 18 var done:[Bool] = [Bool](repeating:false,count:n) 19 for i in 0..<n 20 { 21 if !done[i] 22 { 23 done[i] = true 24 while(p < n && (a[p] != a[i] * 2 || done[p])) 25 { 26 p += 1 27 } 28 if p == n {return false} 29 done[p] = true 30 } 31 } 32 return true 33 } 34 }
1316ms
1 class Solution { 2 func canReorderDoubled(_ A: [Int]) -> Bool { 3 var arr = A.filter { $0 <= 0 }.sorted().reversed() + A.filter { $0 > 0 }.sorted() 4 var unused = [Int: Int]() 5 var used = [Int: Int]() 6 7 for a in arr { 8 unused[a] = (unused[a] ?? 0) + 1 9 } 10 11 for a in arr { 12 if let c = used[a] { 13 if c == 1 { 14 used.removeValue(forKey: a) 15 } else { 16 used[a] = c - 1 17 } 18 } else { 19 unused[a] = unused[a]! - 1 20 if unused[a] == 0 { unused.removeValue(forKey: a) } 21 if let c = unused[a * 2] { 22 if c == 1 { 23 unused.removeValue(forKey: a * 2) 24 } else { 25 unused[a * 2] = c - 1 26 } 27 used[a * 2] = (used[a * 2] ?? 0) + 1 28 } else { 29 return false 30 } 31 } 32 } 33 34 return true 35 } 36 }
1340ms
1 class Solution { 2 func canReorderDoubled(_ A: [Int]) -> Bool { 3 let aSortedPositive = A.filter{ $0 >= 0 }.sorted(by:{ $0 > $1} ) 4 let aSortedNegative = A.filter{ $0 < 0 }.sorted(by:{ $0 < $1} ) 5 let positiveMatches = checkItems(aSortedPositive) 6 let negativeMatches = checkItems(aSortedNegative) 7 8 if positiveMatches && negativeMatches { 9 return true 10 } else { 11 return false 12 } 13 } 14 15 private func checkItems(_ A: [Int]) -> Bool { 16 var pendingNumbers = [Int: Int]() 17 18 for item in A { 19 let theDoubleExists = (pendingNumbers[item * 2] != nil) 20 let theHalfExists = (item%2 == 0 && pendingNumbers[item / 2] != nil) 21 22 if theDoubleExists { 23 var newDoubleCount = pendingNumbers[item * 2]! - 1 24 if newDoubleCount == 0 { 25 pendingNumbers.removeValue(forKey:item * 2) 26 } else { 27 pendingNumbers[item * 2] = newDoubleCount 28 } 29 } else if theHalfExists { 30 var newHalfCount = pendingNumbers[item / 2]! - 1 31 if newHalfCount == 0 { 32 pendingNumbers.removeValue(forKey:item / 2) 33 } else { 34 pendingNumbers[item / 2] = newHalfCount 35 } 36 37 } else { 38 var newCount = (pendingNumbers[item] ?? 0) + 1 39 pendingNumbers[item] = newCount 40 } 41 42 } 43 44 if pendingNumbers.count == 0 { 45 return true 46 } else { 47 return false 48 } 49 } 50 }
1908ms
1 class Solution { 2 func canReorderDoubled(_ A: [Int]) -> Bool { 3 return checkItems(A.sorted(by:{ abs($0) > abs($1)})) 4 } 5 6 private func checkItems(_ A: [Int]) -> Bool { 7 var pendingNumbers = [Int: Int]() 8 9 for item in A { 10 let theDoubleCount = pendingNumbers[item * 2] 11 let theDoubleExists = (theDoubleCount != nil && theDoubleCount != 0) 12 13 if theDoubleExists { 14 pendingNumbers[item * 2] = theDoubleCount! - 1 15 } else { 16 var newCount = (pendingNumbers[item] ?? 0) + 1 17 pendingNumbers[item] = newCount 18 } 19 } 20 21 let remainingNumbers = pendingNumbers.filter{ $0.value != 0 } 22 23 if remainingNumbers.count == 0 { 24 return true 25 } else { 26 return false 27 } 28 } 29 }
2036ms
1 class Solution { 2 func canReorderDoubled(_ A: [Int]) -> Bool { 3 guard A.count > 0 else { return true } 4 var sortedA = A.sorted() 5 var sortedNegA: [Int] = [] 6 var map: [Int: Int] = [:] 7 for a in A { 8 map[a] = (map[a] ?? 0) + 1 9 } 10 11 if sortedA[0] < 0 { 12 var i = 0 13 while i < sortedA.count { 14 if sortedA[i] < 0 { 15 sortedNegA.append(sortedA[i]) 16 sortedA.remove(at: i) 17 } else { 18 break 19 } 20 } 21 } 22 23 sortedNegA = Array(sortedNegA.reversed()) 24 //print(sortedA) 25 //print(sortedNegA) 26 var i = 0 27 while i < sortedNegA.count { 28 defer { i += 1} 29 guard (map[sortedNegA[i]] ?? 0) > 0 else { continue } 30 if (map[sortedNegA[i] * 2] ?? 0) > 0 { 31 map[sortedNegA[i] * 2] = map[sortedNegA[i] * 2]! - 1 32 map[sortedNegA[i]] = map[sortedNegA[i]]! - 1 33 } else { 34 return false 35 } 36 } 37 38 i = 0 39 while i < sortedA.count { 40 defer { i += 1} 41 guard (map[sortedA[i]] ?? 0) > 0 else { continue } 42 if (map[sortedA[i] * 2] ?? 0) > 0 { 43 map[sortedA[i] * 2] = map[sortedA[i] * 2]! - 1 44 map[sortedA[i]] = map[sortedA[i]]! - 1 45 } else { 46 return false 47 } 48 } 49 50 return true 51 } 52 }
