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

标签：def   ace   items   default   代码   ati   pytho   BMI   turn   

题目如下：

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

解题思路：我的方法和 3Sum 题一样，就是先算出A中任意两个数的与值，然后再和A中所有值与操作判断是否为0，耗时3秒多。不管怎么样，至少通过了。

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

代码如下：

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

标签：def   ace   items   default   代码   ati   pytho   BMI   turn   

原文地址：https://www.cnblogs.com/seyjs/p/10356022.html