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


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