[Swift]LeetCode923.三数之和的多种可能 | 3Sum With Multiplicity
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Given an integer array A, and an integer target, return the number of tuples i, j, k such that i < j < kand A[i] + A[j] + A[k] == target.
As the answer can be very large, return it modulo 10^9 + 7.
Example 1:
Input: A = [1,1,2,2,3,3,4,4,5,5], target = 8
Output: 20
Explanation:
Enumerating by the values (A[i], A[j], A[k]):
(1, 2, 5) occurs 8 times;
(1, 3, 4) occurs 8 times;
(2, 2, 4) occurs 2 times;
(2, 3, 3) occurs 2 times.
Example 2:
Input: A = [1,1,2,2,2,2], target = 5
Output: 12
Explanation:
A[i] = 1, A[j] = A[k] = 2 occurs 12 times:
We choose one 1 from [1,1] in 2 ways,
and two 2s from [2,2,2,2] in 6 ways.
Note:
3 <= A.length <= 30000 <= A[i] <= 1000 <= target <= 300
给定一个整数数组 A，以及一个整数 target 作为目标值，返回满足 i < j < k 且 A[i] + A[j] + A[k] == target 的元组 i, j, k 的数量。
由于结果会非常大，请返回 结果除以 10^9 + 7 的余数。
示例 1：
输入：A = [1,1,2,2,3,3,4,4,5,5], target = 8 输出：20 解释： 按值枚举（A[i]，A[j]，A[k]）： (1, 2, 5) 出现 8 次； (1, 3, 4) 出现 8 次； (2, 2, 4) 出现 2 次； (2, 3, 3) 出现 2 次。
示例 2：
输入：A = [1,1,2,2,2,2], target = 5 输出：12 解释： A[i] = 1，A[j] = A[k] = 2 出现 12 次： 我们从 [1,1] 中选择一个 1，有 2 种情况， 从 [2,2,2,2] 中选出两个 2，有 6 种情况。
提示：
3 <= A.length <= 30000 <= A[i] <= 1000 <= target <= 300
80ms
1 struct ModInt { 2 static let MOD = 1000_000_007; 3 4 var val : Int = 0 5 6 init(_ v: Int) { 7 val = v 8 } 9 10 static func *(l: ModInt, r: ModInt) -> ModInt { 11 return ModInt(l.val * r.val % MOD) 12 } 13 static func *(l: ModInt, r: Int) -> ModInt { 14 return ModInt(l.val * r % MOD) 15 } 16 static func +(l: ModInt, r: ModInt) -> ModInt { 17 return ModInt(l.val + r.val % MOD) 18 } 19 static func +(l: ModInt, r: Int) -> ModInt { 20 return ModInt(l.val + r % MOD) 21 } 22 static func -(l: ModInt, r: ModInt) -> ModInt { 23 var res = l.val - r.val 24 if res < 0 { 25 res += MOD 26 } 27 return ModInt(res) 28 } 29 } 30 31 class Solution { 32 33 func threeSumMulti(_ A: [Int], _ target: Int) -> Int { 34 var cnt: [Int:Int] = [:] 35 for v in A { 36 if let c = cnt[v] { 37 cnt[v] = c + 1 38 } else { 39 cnt[v] = 1 40 } 41 } 42 43 var result = ModInt(0) 44 45 let keys = Array(cnt.keys) 46 for ai in 0..<keys.count { 47 let a = keys[ai] 48 for bi in ai..<keys.count { 49 let b = keys[bi] 50 for ci in bi..<keys.count { 51 let c = keys[ci] 52 if a+b+c == target { 53 if a == b && b == c { 54 let n = cnt[a]! 55 result = result + n*(n-1)*(n-2) / 6 56 } else if a == b { 57 let n = cnt[a]! 58 result = result + n*(n-1)*cnt[c]! / 2 59 } else if a == c { 60 let n = cnt[a]! 61 result = result + n*(n-1)*cnt[b]! / 2 62 } else if b == c { 63 let n = cnt[b]! 64 result = result + n*(n-1)*cnt[a]! / 2 65 } else { 66 result = result + cnt[a]!*cnt[b]!*cnt[c]! 67 } 68 } 69 } 70 } 71 } 72 return result.val 73 } 74 }
148ms
1 class Solution { 2 func threeSumMulti(_ A: [Int], _ target: Int) -> Int { 3 guard A.count >= 3 else { 4 return 0 5 } 6 let A = A.sorted() 7 var results = [[Int]]() 8 var dict = Dictionary( A.map { ($0, 1) }, uniquingKeysWith: +) 9 var counter = 0 10 for i in 0...A.count - 3 { 11 if i != 0 && A[i] == A[i-1] { 12 continue 13 } 14 var left = i + 1 15 var right = A.count - 1 16 while left < right { 17 if left != i + 1 && A[left] == A[left - 1] { 18 left += 1 19 continue 20 } 21 if A[i] + A[left] + A[right] == target { 22 let temp = [A[i], A[left], A[right]] 23 results.append(temp) 24 left += 1 25 right -= 1 26 } else if A[i] + A[left] + A[right] > target { 27 right -= 1 28 } else { 29 left += 1 30 } 31 } 32 } 33 print(results) 34 for result in results { 35 if result[0] == result[1] && result[1] == result[2] { 36 let n = dict[result[0]]! 37 counter += n * (n - 1) * (n - 2) / 6 38 39 } else if result[0] == result[1] { 40 41 let n = dict[result[0]]! 42 counter += (n * n / 2 - n / 2) * dict[result[2]]! 43 44 } else if result[1] == result[2] { 45 46 let n = dict[result[2]]! 47 counter += (n * n / 2 - n / 2) * dict[result[0]]! 48 49 } else { 50 counter += dict[result[0]]! * dict[result[1]]! * dict[result[2]]! 51 52 } 53 } 54 return counter % Int(1e9 + 7) 55 } 56 }
308ms
1 class Solution { 2 func threeSumMulti(_ A: [Int], _ target: Int) -> Int { 3 var MOD:Int = 1_000_000_007; 4 var ans:Int = 0; 5 //升序 6 var arr:[Int] = A.sorted(by: <) 7 let len:Int = arr.count 8 for i in 0..<len 9 { 10 //找出i < j < k， 11 //当T = target - arr[i],arr[j] + arr[k] == T成立 . 12 var T:Int = target - arr[i] 13 var j:Int = i + 1, k:Int = len - 1 14 while(j < k) 15 { 16 if arr[j] + arr[k] < T {j += 1} 17 else if arr[j] + arr[k] > T {k -= 1} 18 else if arr[j] != arr[k] 19 { 20 var left:Int = 1, right:Int = 1 21 while(j+1 < k && arr[j] == arr[j+1]) 22 { 23 left += 1 24 j += 1 25 } 26 while(k-1 > j && arr[k] == arr[k-1]) 27 { 28 right += 1 29 k -= 1 30 } 31 ans += left * right 32 ans %= MOD 33 j += 1 34 k -= 1 35 } 36 else 37 { 38 ans += (k-j+1) * (k-j) / 2 39 ans %= MOD 40 break 41 } 42 } 43 } 44 return ans 45 } 46 }
372ms
1 class Solution { 2 func threeSumMulti(_ A: [Int], _ target: Int) -> Int { 3 var MOD:Int = 1_000_000_007; 4 var ans:Int = 0; 5 //升序 6 var arr:[Int] = A.sorted(by: <) 7 let len:Int = arr.count 8 for i in 0..<len 9 { 10 //找出i < j < k， 11 //当T = target - arr[i],arr[j] + arr[k] == T成立 . 12 var T:Int = target - arr[i] 13 var j:Int = i + 1, k:Int = len - 1 14 while(j < k) 15 { 16 if arr[j] + arr[k] < T {j += 1} 17 else if arr[j] + arr[k] > T {k -= 1} 18 else if arr[j] != arr[k] 19 { 20 var left:Int = 1, right:Int = 1 21 while(j+1 < k && arr[j] == arr[j+1]) 22 { 23 left += 1 24 j += 1 25 } 26 while(k-1 > j && arr[k] == arr[k-1]) 27 { 28 right += 1 29 k -= 1 30 } 31 ans += left * right 32 ans %= MOD 33 j += 1 34 k -= 1 35 } 36 else 37 { 38 ans += (k-j+1) * (k-j) / 2 39 ans %= MOD 40 break 41 } 42 } 43 } 44 return ans 45 } 46 }
