快手：笔试题（20190916）

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#快手 #笔试题

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本文介绍了一种用于验证和区分IPv4与IPv6地址的算法，并提供了小灵通键盘字母组合的解决方案，同时探讨了集合划分及最长等差数列的计算方法。

验证IP地址

分类别进行筛选，注意IPv4是以.为分割符，而IPv6是以:为分隔符。

def confine_ip_address(seq):
    if not seq:
        return 'Neither'
    if '.' in seq:
        seq = seq.split('.')
        # print('.', seq)
        if len(seq) == 4:
            return find_ipv4(seq)
    if ':' in seq:
        seq = seq.split(':')
        # print(':', seq)
        if len(seq) == 8:
            return find_ipv6(seq)
    if len(seq) != 4 or len(seq) != 6:
        return 'Neither'


def find_ipv4(seq):
    for su in seq:
        # print(su)
        if len(su) > 1 and su.startswith('0'):
            return 'Neither'
        if su < '0' or su > '255':
            return 'Neither'
    return 'IPv4'


def find_ipv6(seq):
    values = set(list(map(str, range(10))) + list('abcedfABCDEF'))
    for su in seq:
        # print(su)
        if len(su) > 1 and set(su) == {'0'}:
            return 'Neither'
        for s in su:
            if s not in values:
                return 'Neither'
    return 'IPv6'


if __name__ == '__main__':
    seq = input().strip()
    res = confine_ip_address(seq)
    print(res)

'''
172.16.254.1

2001:0db8:85a3:0:0:8A2E:0370:7334

256.256.256.256
'''

字母组合

小灵通的键盘，有意思。每次只对两个数字分别代表的意思进行组合，依次遍历下去。

def alpha_combine(seq):
    phone = ['abc', 'def', 'ghi', 'jkl', 'mno', 'pqrs', 'tuv', 'wxyz']
    values = {nu: pu for nu, pu in zip(range(2, 10), phone)}
    pre = values[seq[0]]
    for su in seq[1:]:
        left, right = pre, values[su]
        current = []
        for lu in left:
            for ru in right:
                current.append(''.join([lu, ru]))
        pre = current
    return pre


if __name__ == '__main__':
    seq = list(map(int, list(input().strip())))
    seq = [su for su in seq if su != 1]
    res = alpha_combine(seq)
    print('[', end='')
    print(', '.join(res), end='')

'''
23

92

458
'''

集合划分问题

懒人模式，暴力求解。通过率为57.13%，栈溢出，很正常的现象。

from itertools import permutations


def split_set_problem(seq, n):
    index = []
    for nu in range(1, n):
        index.extend(list(permutations(range(n), nu)))
    # print(index)
    res = float('inf')
    for idx in index:
        left = [seq[si] for si in idx]
        right = [seq[ri] for ri in range(n) if ri not in idx]
        current = abs(sum(left)-sum(right))
        if current < res:
            res = current
    return res


if __name__ == '__main__':
    n = int(input().strip())
    seq = list(map(int, input().strip().split()))
    res = split_set_problem(seq, n)
    print(res)

'''
5
2 4 5 6 9

4
1 1 1 999
'''

最长等差数列

再次懒人模式，暴力求解。通过率80%，超时，太正常了。

def longest_array_for_equal(seq):
    if not seq:
        return 0
    seq.sort()
    res = 0
    left, right = 0, seq[-1]-seq[0]
    for si in range(len(seq)):
        idx, cur = si, 0
        su = seq[si]
        cur = su
        for iu in range(left, right+1):
            count = 1
            idx = si
            while True:
                # print(cur, count, idx)
                if cur+iu in seq[idx+1:]:
                    idx = seq[idx+1:].index(cur + iu)
                    cur += iu
                    count += 1
                else:
                    break
            if count > res:
                res = count
    return res


if __name__ == '__main__':
    n = int(input().strip())
    seq = list(map(int, input().strip().split()))
    res = longest_array_for_equal(seq)
    print(res)

'''
5
1 4 2 5 3

5
1 4 7 2 6
'''