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 <= A.length <= 1000
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