[Swift]LeetCode982. 按位与为零的三元组 | Triples with Bitwise AND Equal To Zero

原文地址：https://www.cnblogs.com/strengthen/p/10326155.html 

Given an array of integers A, find the number of triples of indices (i, j, k) such that:

• 0 <= i < A.length
• 0 <= j < A.length
• 0 <= k < A.length
• A[i] & A[j] & A[k] == 0, where & represents the bitwise-AND operator. 

Example 1:

Input: [2,1,3]
Output: 12
Explanation: We could choose the following i, j, k triples:
(i=0, j=0, k=1) : 2 & 2 & 1
(i=0, j=1, k=0) : 2 & 1 & 2
(i=0, j=1, k=1) : 2 & 1 & 1
(i=0, j=1, k=2) : 2 & 1 & 3
(i=0, j=2, k=1) : 2 & 3 & 1
(i=1, j=0, k=0) : 1 & 2 & 2
(i=1, j=0, k=1) : 1 & 2 & 1
(i=1, j=0, k=2) : 1 & 2 & 3
(i=1, j=1, k=0) : 1 & 1 & 2
(i=1, j=2, k=0) : 1 & 3 & 2
(i=2, j=0, k=1) : 3 & 2 & 1
(i=2, j=1, k=0) : 3 & 1 & 2 

Note:

1. 1 <= A.length <= 1000
2. 0 <= A[i] < 2^16

---

给定一个整数数组 A，找出索引为 (i, j, k) 的三元组，使得：

• 0 <= i < A.length
• 0 <= j < A.length
• 0 <= k < A.length
• A[i] & A[j] & A[k] == 0，其中 & 表示按位与（AND）操作符。 

示例：

输入：[2,1,3]
输出：12
解释：我们可以选出如下 i, j, k 三元组：
(i=0, j=0, k=1) : 2 & 2 & 1
(i=0, j=1, k=0) : 2 & 1 & 2
(i=0, j=1, k=1) : 2 & 1 & 1
(i=0, j=1, k=2) : 2 & 1 & 3
(i=0, j=2, k=1) : 2 & 3 & 1
(i=1, j=0, k=0) : 1 & 2 & 2
(i=1, j=0, k=1) : 1 & 2 & 1
(i=1, j=0, k=2) : 1 & 2 & 3
(i=1, j=1, k=0) : 1 & 1 & 2
(i=1, j=2, k=0) : 1 & 3 & 2
(i=2, j=0, k=1) : 3 & 2 & 1
(i=2, j=1, k=0) : 3 & 1 & 2 

提示：

1. 1 <= A.length <= 1000
2. 0 <= A[i] < 2048

---

9888ms

class Solution {
    func countTriplets(_ A: [Int]) -> Int {
        var cnt:[Int] = [Int](repeating:0,count:65536)
        for i in 0..<65536
        {
            for n in A
            {
                if (i&n)==0
                {
                    cnt[i] += 1
                }
            }
        }
        
        var res:Int = 0
        for a in A
        {
            for b in A
            {
                res += cnt[a&b]
            }
        }
        return res
    }
}