題目如下：

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

解題思路：我的方法和 3Sum 題一樣，就是先算出A中任意兩個數的與值，然後再和A中所有值與操作判斷是否為0，耗時3秒多。不管怎麽樣，至少通過了。

Runtime: 3772 ms, faster than 39.02% of Python online submissions for Triples with Bitwise AND Equal To Zero.

代碼如下：

class Solution(object):
    def countTriplets(self, A):
        
 
"""
        :type A: List[int]
        :rtype: int
        """
        dic = {}
        for i in A:
            for j in A:
                v = i & j
                dic[v] = dic.setdefault(v,0) + 1
        res = 0
        for i in A:
            for k,v in dic.iteritems():
                if i & k == 0:
                    res 
 
+= v
        return res

【leetcode】982. Triples with Bitwise AND Equal To Zero