• [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:

    1. 3 <= A.length <= 3000
    2. 0 <= A[i] <= 100
    3. 0 <= 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:

    1. 3 <= A.length <= 3000
    2. 0 <= A[i] <= 100
    3. 0 <= 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 }
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  • 原文地址:https://www.cnblogs.com/strengthen/p/9830546.html
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