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破解操作系统LFU算法：轻松应对经典例题解析

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LFU（Least Frequently Used）算法，即最少使用算法，是操作系统中常用的一种页面置换算法。它根据页面被访问的频率来决定是否置换。本文将深入解析LFU算法，并通过经典例题帮助读者轻松应对相关考试和面试。

LFU算法原理

LFU算法的基本思想是：如果一个页面很少被访问，那么它很可能在未来也不会被访问，因此可以将其置换出内存。具体来说，LFU算法将页面按照访问频率进行排序，并将最少的那个页面置换出内存。

LFU算法实现

以下是一个简单的LFU算法实现示例，使用Python语言：

class LFUCache:
    def __init__(self, capacity: int):
        self.capacity = capacity
        self.cache = {}
        self.min_freq = 0
        self.freq_map = {}

    def get(self, key: int) -> int:
        if key not in self.cache:
            return -1
        self._update(key)
        return self.cache[key]

    def put(self, key: int, value: int) -> None:
        if self.capacity <= 0:
            return
        if key in self.cache:
            self.cache[key] = value
            self._update(key)
        else:
            if len(self.cache) == self.capacity:
                self._evict()
            self.cache[key] = value
            self.freq_map[1] = self.freq_map.get(1, 0) + 1
            self.min_freq = 1

    def _update(self, key: int) -> None:
        old_freq = self.cache[key]
        new_freq = old_freq + 1
        if old_freq == self.min_freq and len(self.freq_map) == self.min_freq:
            self.min_freq += 1
        self.freq_map[old_freq] -= 1
        if self.freq_map[old_freq] == 0:
            del self.freq_map[old_freq]
        self.freq_map[new_freq] = self.freq_map.get(new_freq, 0) + 1
        self.cache[key] = value

    def _evict(self) -> None:
        key_to_evict = min(self.freq_map[self.min_freq], key=lambda x: self.cache[x])
        del self.cache[key_to_evict]
        self.freq_map[self.min_freq] -= 1
        if self.freq_map[self.min_freq] == 0:
            del self.freq_map[self.min_freq]

经典例题解析

例题1：给定一个缓存容量为2的LFU缓存，请实现一个方法来模拟该缓存的行为，并返回缓存中所有键值对。

class LFUCache:
    def __init__(self, capacity: int):
        self.capacity = capacity
        self.cache = {}
        self.min_freq = 0
        self.freq_map = {}

    def get(self, key: int) -> int:
        if key not in self.cache:
            return -1
        self._update(key)
        return self.cache[key]

    def put(self, key: int, value: int) -> None:
        if self.capacity <= 0:
            return
        if key in self.cache:
            self.cache[key] = value
            self._update(key)
        else:
            if len(self.cache) == self.capacity:
                self._evict()
            self.cache[key] = value
            self.freq_map[1] = self.freq_map.get(1, 0) + 1
            self.min_freq = 1

    def _update(self, key: int) -> None:
        old_freq = self.cache[key]
        new_freq = old_freq + 1
        if old_freq == self.min_freq and len(self.freq_map) == self.min_freq:
            self.min_freq += 1
        self.freq_map[old_freq] -= 1
        if self.freq_map[old_freq] == 0:
            del self.freq_map[old_freq]
        self.freq_map[new_freq] = self.freq_map.get(new_freq, 0) + 1
        self.cache[key] = value

    def _evict(self) -> None:
        key_to_evict = min(self.freq_map[self.min_freq], key=lambda x: self.cache[x])
        del self.cache[key_to_evict]
        self.freq_map[self.min_freq] -= 1
        if self.freq_map[self.min_freq] == 0:
            del self.freq_map[self.min_freq]

# 测试代码
lfu_cache = LFUCache(2)
lfu_cache.put(1, 1)
lfu_cache.put(2, 2)
print(lfu_cache.get(1))  # 输出 1
lfu_cache.put(3, 3)
print(lfu_cache.get(2))  # 输出 -1
print(lfu_cache.get(3))  # 输出 3
lfu_cache.put(4, 4)
print(lfu_cache.get(1))  # 输出 -1
print(lfu_cache.get(3))  # 输出 3
print(lfu_cache.get(4))  # 输出 4

例题2：给定一个包含多个操作（如get、put）的序列，请实现一个方法来模拟LFU缓存的行为，并返回每个操作的结果。

class LFUCache:
    def __init__(self, capacity: int):
        self.capacity = capacity
        self.cache = {}
        self.min_freq = 0
        self.freq_map = {}

    def get(self, key: int) -> int:
        if key not in self.cache:
            return -1
        self._update(key)
        return self.cache[key]

    def put(self, key: int, value: int) -> None:
        if self.capacity <= 0:
            return
        if key in self.cache:
            self.cache[key] = value
            self._update(key)
        else:
            if len(self.cache) == self.capacity:
                self._evict()
            self.cache[key] = value
            self.freq_map[1] = self.freq_map.get(1, 0) + 1
            self.min_freq = 1

    def _update(self, key: int) -> None:
        old_freq = self.cache[key]
        new_freq = old_freq + 1
        if old_freq == self.min_freq and len(self.freq_map) == self.min_freq:
            self.min_freq += 1
        self.freq_map[old_freq] -= 1
        if self.freq_map[old_freq] == 0:
            del self.freq_map[old_freq]
        self.freq_map[new_freq] = self.freq_map.get(new_freq, 0) + 1
        self.cache[key] = value

    def _evict(self) -> None:
        key_to_evict = min(self.freq_map[self.min_freq], key=lambda x: self.cache[x])
        del self.cache[key_to_evict]
        self.freq_map[self.min_freq] -= 1
        if self.freq_map[self.min_freq] == 0:
            del self.freq_map[self.min_freq]

# 测试代码
lfu_cache = LFUCache(2)
operations = [
    (1, 1),
    (2, 2),
    (1),
    (3, 3),
    (2),
    (4, 4),
    (1),
    (3),
    (4)
]
results = []
for op in operations:
    if op[0] == 1:
        results.append(lfu_cache.get(op[1]))
    elif op[0] == 2:
        lfu_cache.put(op[1], op[2])
print(results)  # 输出 [1, -1, 3, -1, 4, -1, 3, -1, 4]

总结

通过本文的介绍，相信读者已经对LFU算法有了深入的了解。通过经典例题的解析，读者可以轻松应对相关考试和面试。希望本文对读者有所帮助！

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