当今顶尖科技公司的数据团队面试中,算法能力考核占据决定性地位。与常规算法面试不同,大数据场景下的题目往往具有三个显著特征:
下面我们通过六大核心题型,深入剖析大数据算法面试的解题方法论。
问题描述:给定包含10亿个整数的文件,找出出现频率最高的100个数。
哈希表+堆排序方案:
import heapq
from collections import defaultdict
def top_k_frequent(nums, k):
# 统计频率 O(n)
freq_map = defaultdict(int)
for num in nums:
freq_map[num] += 1
# 维护最小堆 O(nlogk)
heap = []
for num, freq in freq_map.items():
if len(heap) < k:
heapq.heappush(heap, (freq, num))
else:
if freq > heap[0][0]:
heapq.heappop(heap)
heapq.heappush(heap, (freq, num))
return [num for (freq, num) in heap]
复杂度分析:
MapReduce实现方案:
# Mapper阶段
def mapper(chunk):
local_count = {}
for num in chunk:
local_count[num] = local_count.get(num, 0) + 1
yield from local_count.items()
# Reducer阶段
def reducer(results, k):
global_count = {}
for num, cnt in results:
global_count[num] = global_count.get(num, 0) + cnt
# 使用堆获取TopK
return heapq.nlargest(k, global_count.items(), key=lambda x: x[1])
优化要点:
问题描述:从数据流中随机选取k个元素,保证每个元素被选中的概率相同。
算法实现:
import random
def reservoir_sampling(stream, k):
reservoir = []
for i, item in enumerate(stream):
if i < k:
reservoir.append(item)
else:
j = random.randint(0, i)
if j < k:
reservoir[j] = item
return reservoir
数学证明:
HyperLogLog实现:
import hashlib
import math
class HyperLogLog:
def __init__(self, precision=12):
self.p = precision
self.m = 1 << precision
self.registers = [0] * self.m
def add(self, element):
hashed = int(hashlib.md5(element.encode()).hexdigest(), 16)
index = hashed & (self.m - 1)
w = hashed >> self.p
self.registers[index] = max(self.registers[index],
(w & -w).bit_length())
def count(self):
harmonic_mean = sum(2**-r for r in self.registers)
estimate = self.m * self.m * 0.7213 / harmonic_mean
return int(estimate)
应用场景:
import hashlib
class ConsistentHash:
def __init__(self, nodes, replica=3):
self.replica = replica
self.ring = {}
for node in nodes:
for i in range(replica):
key = self._hash(f"{node}:{i}")
self.ring[key] = node
self.sorted_keys = sorted(self.ring.keys())
def _hash(self, key):
return int(hashlib.md5(key.encode()).hexdigest(), 16)
def get_node(self, key):
if not self.ring:
return None
hash_key = self._hash(key)
idx = bisect.bisect(self.sorted_keys, hash_key) % len(self.sorted_keys)
return self.ring[self.sorted_keys[idx]]
优化方向:
import mmh3
from bitarray import bitarray
class BloomFilter:
def __init__(self, size, hash_num):
self.size = size
self.hash_num = hash_num
self.bit_array = bitarray(size)
self.bit_array.setall(0)
def add(self, item):
for seed in range(self.hash_num):
index = mmh3.hash(item, seed) % self.size
self.bit_array[index] = 1
def contains(self, item):
for seed in range(self.hash_num):
index = mmh3.hash(item, seed) % self.size
if not self.bit_array[index]:
return False
return True
参数计算:
class BitMap:
def __init__(self, max_num):
self._size = (max_num + 7) // 8
self.bitmap = bytearray(self._size)
def set(self, num):
index = num // 8
offset = num % 8
self.bitmap[index] |= 1 << offset
def get(self, num):
index = num // 8
offset = num % 8
return (self.bitmap[index] & (1 << offset)) != 0
应用场景:
掌握大数据算法需要建立三层能力:
建议按照以下路径系统提升:
记住,优秀的工程师不仅要知道如何解决问题,更要理解为什么选择这种解决方案。这正是在大数据算法面试中脱颖而出的关键。
原创声明:本文系作者授权腾讯云开发者社区发表,未经许可,不得转载。
如有侵权,请联系 cloudcommunity@tencent.com 删除。
原创声明:本文系作者授权腾讯云开发者社区发表,未经许可,不得转载。
如有侵权,请联系 cloudcommunity@tencent.com 删除。