说明

        如果需要用到这些知识却没有掌握，则会让人感到沮丧，也可能导致面试被拒。无论是花几天时间“突击”，还是利用零碎的时间持续学习，在数据结构上下点功夫都是值得的。那么Python 中有哪些数据结构呢？列表、字典、集合，还有……栈？Python 有栈吗？本系列文章将给出详细拼图。
 

9章：Advanced Linked Lists

之前曾经介绍过单链表，一个链表节点只有data和next字段，本章介绍高级的链表。

Doubly Linked List，双链表，每个节点多了个prev指向前一个节点。双链表可以用来编写文本编辑器的buffer。

class DListNode:def __init__(self, data):self.data = dataself.prev = Noneself.next = Nonedef revTraversa(tail):curNode = tailwhile cruNode is not None:print(curNode.data)curNode = curNode.prevdef search_sorted_doubly_linked_list(head, tail, probe, target):""" probing technique探查法，改进直接遍历，不过最坏时间复杂度仍是O(n)searching a sorted doubly linked list using the probing techniqueArgs:head (DListNode obj)tail (DListNode obj)probe (DListNode or None)target (DListNode.data): data to search"""if head is None:    # make sure list is not emptyreturn Falseif probe is None:    # if probe is null, initialize it to first nodeprobe = head# if the target comes before the probe node, we traverse backward, otherwise# traverse forwardif target < probe.data:while probe is not None and target <= probe.data:if target == probe.dta:return Trueelse:probe = probe.prevelse:while probe is not None and target >= probe.data:if target == probe.data:return Trueelse:probe = probe.nextreturn Falsedef insert_node_into_ordered_doubly_linekd_list(value):""" 最好画个图看，链表操作很容易绕晕，注意赋值顺序"""newnode = DListNode(value)if head is None:    # empty listhead = newnodetail = headelif value < head.data:    # insert before headnewnode.next = headhead.prev = newnodehead = newnodeelif value > tail.data:    # insert after tailnewnode.prev = tailtail.next = newnodetail = newnodeelse:    # insert into middlenode = headwhile node is not None and node.data < value:node = node.nextnewnode.next = nodenewnode.prev = node.prevnode.prev.next = newnodenode.prev = newnode

循环链表

def travrseCircularList(listRef):curNode = listRefdone = listRef is Nonewhile not None:curNode = curNode.nextprint(curNode.data)done = curNode is listRef   # 回到遍历起始点def searchCircularList(listRef, target):curNode = listRefdone = listRef is Nonewhile not done:curNode = curNode.nextif curNode.data == target:return Trueelse:done = curNode is listRef or curNode.data > targetreturn Falsedef add_newnode_into_ordered_circular_linked_list(listRef, value):""" 插入并维持顺序1.插入空链表；2.插入头部；3.插入尾部；4.按顺序插入中间"""newnode = ListNode(value)if listRef is None:    # empty listlistRef = newnodenewnode.next = newnodeelif value < listRef.next.data:    # insert in frontnewnode.next = listRef.nextlistRef.next = newnodeelif value > listRef.data:    # insert in backnewnode.next = listRef.nextlistRef.next = newnodelistRef = newnodeelse:    # insert in the middlepreNode = NonecurNode = listRefdone = listRef is Nonewhile not done:preNode = curNodepreNode = curNode.nextdone = curNode is listRef or curNode.data > valuenewnode.next = curNodepreNode.next = newnode

利用循环双端链表我们可以实现一个经典的缓存失效算法，lru：

# -*- coding: utf-8 -*-class Node(object):def __init__(self, prev=None, next=None, key=None, value=None):self.prev, self.next, self.key, self.value = prev, next, key, valueclass CircularDoubleLinkedList(object):def __init__(self):node = Node()node.prev, node.next = node, nodeself.rootnode = nodedef headnode(self):return self.rootnode.nextdef tailnode(self):return self.rootnode.prevdef remove(self, node):if node is self.rootnode:returnelse:node.prev.next = node.nextnode.next.prev = node.prevdef append(self, node):tailnode = self.tailnode()tailnode.next = nodenode.next = self.rootnodeself.rootnode.prev = nodeclass LRUCache(object):def __init__(self, maxsize=16):self.maxsize = maxsizeself.cache = {}self.access = CircularDoubleLinkedList()self.isfull = len(self.cache) >= self.maxsizedef __call__(self, func):def wrapper(n):cachenode = self.cache.get(n)if cachenode is not None:  # hitself.access.remove(cachenode)self.access.append(cachenode)return cachenode.valueelse:  # missvalue = func(n)if not self.isfull:tailnode = self.access.tailnode()newnode = Node(tailnode, self.access.rootnode, n, value)self.access.append(newnode)self.cache[n] = newnodeself.isfull = len(self.cache) >= self.maxsizereturn valueelse:  # fulllru_node = self.access.headnode()del self.cache[lru_node.key]self.access.remove(lru_node)tailnode = self.access.tailnode()newnode = Node(tailnode, self.access.rootnode, n, value)self.access.append(newnode)self.cache[n] = newnodereturn valuereturn wrapper@LRUCache()
def fib(n):if n <= 2:return 1else:return fib(n - 1) + fib(n - 2)for i in range(1, 35):print(fib(i))

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10章：Recursion

Recursion is a process for solving problems by subdividing a larger problem into smaller cases of the problem itself and then solving the smaller, more trivial parts.

递归函数：调用自己的函数

# 递归函数：调用自己的函数，看一个最简单的递归函数，倒序打印一个数
def printRev(n):if n > 0:print(n)printRev(n-1)printRev(3)    # 从10输出到1# 稍微改一下，print放在最后就得到了正序打印的函数
def printInOrder(n):if n > 0:printInOrder(n-1)print(n)    # 之所以最小的先打印是因为函数一直递归到n==1时候的最深栈，此时不再# 递归，开始执行print语句，这时候n==1，之后每跳出一层栈，打印更大的值printInOrder(3)    # 正序输出

Properties of Recursion: 使用stack解决的问题都能用递归解决

• A recursive solution must contain a base case; 递归出口，代表最小子问题(n == 0退出打印)
• A recursive solution must contain a recursive case; 可以分解的子问题
• A recursive solution must make progress toward the base case. 递减n使得n像递归出口靠近

Tail Recursion: occurs when a function includes a single recursive call as the last statement of the function. In this case, a stack is not needed to store values to te used upon the return of the recursive call and thus a solution can be implemented using a iterative loop instead.

# Recursive Binary Searchdef recBinarySearch(target, theSeq, first, last):# 你可以写写单元测试来验证这个函数的正确性if first > last:    # 递归出口1return Falseelse:mid = (first + last) // 2if theSeq[mid] == target:return True    # 递归出口2elif theSeq[mid] > target:return recBinarySearch(target, theSeq, first, mid - 1)else:return recBinarySearch(target, theSeq, mid + 1, last)

---

11章：Hash Tables

基于比较的搜索（线性搜索，有序数组的二分搜索）最好的时间复杂度只能达到O(logn)，利用hash可以实现O(1)查找，python内置dict的实现方式就是hash，你会发现dict的key必须要是实现了 __hash__ 和 __eq__ 方法的。

Hashing: hashing is the process of mapping a search a key to a limited range of array indeices with the goal of providing direct access to the keys.

hash方法有个hash函数用来给key计算一个hash值，作为数组下标，放到该下标对应的槽中。当不同key根据hash函数计算得到的下标相同时，就出现了冲突。解决冲突有很多方式，比如让每个槽成为链表，每次冲突以后放到该槽链表的尾部，但是查询时间就会退化，不再是O(1)。还有一种探查方式，当key的槽冲突时候，就会根据一种计算方式去寻找下一个空的槽存放，探查方式有线性探查，二次方探查法等，cpython解释器使用的是二次方探查法。还有一个问题就是当python使用的槽数量大于预分配的2/3时候，会重新分配内存并拷贝以前的数据，所以有时候dict的add操作代价还是比较高的，牺牲空间但是可以始终保证O(1)的查询效率。如果有大量的数据，建议还是使用bloomfilter或者redis提供的HyperLogLog。

如果你感兴趣，可以看看这篇文章，介绍c解释器如何实现的python dict对象：Python dictionary implementation。我们使用Python来实现一个类似的hash结构。

import ctypesclass Array:  # 第二章曾经定义过的ADT，这里当做HashMap的槽数组使用def __init__(self, size):assert size > 0, 'array size must be > 0'self._size = sizePyArrayType = ctypes.py_object * sizeself._elements = PyArrayType()self.clear(None)def __len__(self):return self._sizedef __getitem__(self, index):assert index >= 0 and index < len(self), 'out of range'return self._elements[index]def __setitem__(self, index, value):assert index >= 0 and index < len(self), 'out of range'self._elements[index] = valuedef clear(self, value):""" 设置每个元素为value """for i in range(len(self)):self._elements[i] = valuedef __iter__(self):return _ArrayIterator(self._elements)class _ArrayIterator:def __init__(self, items):self._items = itemsself._idx = 0def __iter__(self):return selfdef __next__(self):if self._idx < len(self._items):val = self._items[self._idx]self._idx += 1return valelse:raise StopIterationclass HashMap:""" HashMap ADT实现，类似于python内置的dict一个槽有三种状态：1.从未使用 HashMap.UNUSED。此槽没有被使用和冲突过，查找时只要找到UNUSEd就不用再继续探查了2.使用过但是remove了，此时是 HashMap.EMPTY，该探查点后边的元素扔可能是有key3.槽正在使用 _MapEntry节点"""class _MapEntry:    # 槽里存储的数据def __init__(self, key, value):self.key = keyself.value = valueUNUSED = None    # 没被使用过的槽，作为该类变量的一个单例，下边都是is 判断EMPTY = _MapEntry(None, None)     # 使用过但是被删除的槽def __init__(self):self._table = Array(7)    # 初始化7个槽self._count = 0# 超过2/3空间被使用就重新分配，load factor = 2/3self._maxCount = len(self._table) - len(self._table) // 3def __len__(self):return self._countdef __contains__(self, key):slot = self._findSlot(key, False)return slot is not Nonedef add(self, key, value):if key in self:    # 覆盖原有valueslot = self._findSlot(key, False)self._table[slot].value = valuereturn Falseelse:slot = self._findSlot(key, True)self._table[slot] = HashMap._MapEntry(key, value)self._count += 1if self._count == self._maxCount:    # 超过2/3使用就rehashself._rehash()return Truedef valueOf(self, key):slot = self._findSlot(key, False)assert slot is not None, 'Invalid map key'return self._table[slot].valuedef remove(self, key):""" remove操作把槽置为EMPTY"""assert key in self, 'Key error %s' % keyslot = self._findSlot(key, forInsert=False)value = self._table[slot].valueself._count -= 1self._table[slot] = HashMap.EMPTYreturn valuedef __iter__(self):return _HashMapIteraotr(self._table)def _slot_can_insert(self, slot):return (self._table[slot] is HashMap.EMPTY orself._table[slot] is HashMap.UNUSED)def _findSlot(self, key, forInsert=False):""" 注意原书有错误，代码根本不能运行，这里我自己改写的Args:forInsert (bool): if the search is for an insertionReturns:slot or None"""slot = self._hash1(key)step = self._hash2(key)_len = len(self._table)if not forInsert:    # 查找是否存在keywhile self._table[slot] is not HashMap.UNUSED:# 如果一个槽是UNUSED，直接跳出if self._table[slot] is HashMap.EMPTY:slot = (slot + step) % _lencontinueelif self._table[slot].key == key:return slotslot = (slot + step) % _lenreturn Noneelse:    # 为了插入keywhile not self._slot_can_insert(slot):    # 循环直到找到一个可以插入的槽slot = (slot + step) % _lenreturn slotdef _rehash(self):    # 当前使用槽数量大于2/3时候重新创建新的tableorigTable = self._tablenewSize = len(self._table) * 2 + 1    # 原来的2*n+1倍self._table = Array(newSize)self._count = 0self._maxCount = newSize - newSize // 3# 将原来的key value添加到新的tablefor entry in origTable:if entry is not HashMap.UNUSED and entry is not HashMap.EMPTY:slot = self._findSlot(entry.key, True)self._table[slot] = entryself._count += 1def _hash1(self, key):""" 计算key的hash值"""return abs(hash(key)) % len(self._table)def _hash2(self, key):""" key冲突时候用来计算新槽的位置"""return 1 + abs(hash(key)) % (len(self._table)-2)class _HashMapIteraotr:def __init__(self, array):self._array = arrayself._idx = 0def __iter__(self):return selfdef __next__(self):if self._idx < len(self._array):if self._array[self._idx] is not None and self._array[self._idx].key is not None:key = self._array[self._idx].keyself._idx += 1return keyelse:self._idx += 1else:raise StopIterationdef print_h(h):for idx, i in enumerate(h):print(idx, i)print('\n')def test_HashMap():""" 一些简单的单元测试，不过测试用例覆盖不是很全面 """h = HashMap()assert len(h) == 0h.add('a', 'a')assert h.valueOf('a') == 'a'assert len(h) == 1a_v = h.remove('a')assert a_v == 'a'assert len(h) == 0h.add('a', 'a')h.add('b', 'b')assert len(h) == 2assert h.valueOf('b') == 'b'b_v = h.remove('b')assert b_v == 'b'assert len(h) == 1h.remove('a')assert len(h) == 0n = 10for i in range(n):h.add(str(i), i)assert len(h) == nprint_h(h)for i in range(n):assert str(i) in hfor i in range(n):h.remove(str(i))assert len(h) == 0

---

12章: Advanced Sorting

第5章介绍了基本的排序算法，本章介绍高级排序算法。

归并排序(mergesort): 分治法

def merge_sorted_list(listA, listB):""" 归并两个有序数组，O(max(m, n)) ,m和n是数组长度"""print('merge left right list', listA, listB, end='')new_list = list()a = b = 0while a < len(listA) and b < len(listB):if listA[a] < listB[b]:new_list.append(listA[a])a += 1else:new_list.append(listB[b])b += 1while a < len(listA):new_list.append(listA[a])a += 1while b < len(listB):new_list.append(listB[b])b += 1print(' ->', new_list)return new_listdef mergesort(theList):""" O(nlogn)，log层调用，每层n次操作mergesort: divided and conquer 分治1. 把原数组分解成越来越小的子数组2. 合并子数组来创建一个有序数组"""print(theList)    # 我把关键步骤打出来了，你可以运行下看看整个过程if len(theList) <= 1:    # 递归出口return theListelse:mid = len(theList) // 2# 递归分解左右两边数组left_half = mergesort(theList[:mid])right_half = mergesort(theList[mid:])# 合并两边的有序子数组newList = merge_sorted_list(left_half, right_half)return newList""" 这是我调用一次打出来的排序过程
[10, 9, 8, 7, 6, 5, 4, 3, 2, 1]
[10, 9, 8, 7, 6]
[10, 9]
[10]
[9]
merge left right list [10] [9] -> [9, 10]
[8, 7, 6]
[8]
[7, 6]
[7]
[6]
merge left right list [7] [6] -> [6, 7]
merge left right list [8] [6, 7] -> [6, 7, 8]
merge left right list [9, 10] [6, 7, 8] -> [6, 7, 8, 9, 10]
[5, 4, 3, 2, 1]
[5, 4]
[5]
[4]
merge left right list [5] [4] -> [4, 5]
[3, 2, 1]
[3]
[2, 1]
[2]
[1]
merge left right list [2] [1] -> [1, 2]
merge left right list [3] [1, 2] -> [1, 2, 3]
merge left right list [4, 5] [1, 2, 3] -> [1, 2, 3, 4, 5]
"""

快速排序

def quicksort(theSeq, first, last):    # average: O(nlog(n))"""quicksort :也是分而治之，但是和归并排序不同的是，采用选定主元（pivot）而不是从中间进行数组划分1. 第一步选定pivot用来划分数组，pivot左边元素都比它小，右边元素都大于等于它2. 对划分的左右两边数组递归，直到递归出口（数组元素数目小于2）3. 对pivot和左右划分的数组合并成一个有序数组"""if first < last:pos = partitionSeq(theSeq, first, last)# 对划分的子数组递归操作quicksort(theSeq, first, pos - 1)quicksort(theSeq, pos + 1, last)def partitionSeq(theSeq, first, last):""" 快排中的划分操作，把比pivot小的挪到左边，比pivot大的挪到右边"""pivot = theSeq[first]print('before partitionSeq', theSeq)left = first + 1right = lastwhile True:# 找到第一个比pivot大的while left <= right and theSeq[left] < pivot:left += 1# 从右边开始找到比pivot小的while right >= left and theSeq[right] >= pivot:right -= 1if right < left:breakelse:theSeq[left], theSeq[right] = theSeq[right], theSeq[left]# 把pivot放到合适的位置theSeq[first], theSeq[right] = theSeq[right], theSeq[first]print('after partitionSeq {}: {}\t'.format(theSeq, pivot))return right    # 返回pivot的位置def test_partitionSeq():l = [0,1,2,3,4]assert partitionSeq(l, 0, len(l)-1) == 0l = [4,3,2,1,0]assert partitionSeq(l, 0, len(l)-1) == 4l = [2,3,0,1,4]assert partitionSeq(l, 0, len(l)-1) == 2test_partitionSeq()def test_quicksort():def _is_sorted(seq):for i in range(len(seq)-1):if seq[i] > seq[i+1]:return Falsereturn Truefrom random import randintfor i in range(100):_len = randint(1, 100)to_sort = []for i in range(_len):to_sort.append(randint(0, 100))quicksort(to_sort, 0, len(to_sort)-1)    # 注意这里用了原地排序，直接更改了数组print(to_sort)assert _is_sorted(to_sort)test_quicksort()

利用快排中的partitionSeq操作，我们还能实现另一个算法，nth_element，快速查找一个无序数组中的第k大元素

def nth_element(seq, beg, end, k):if beg == end:return seq[beg]pivot_index = partitionSeq(seq, beg, end)if pivot_index == k:return seq[k]elif pivot_index > k:return nth_element(seq, beg, pivot_index-1, k)else:return nth_element(seq, pivot_index+1, end, k)def test_nth_element():from random import shufflen = 10l = list(range(n))shuffle(l)print(l)for i in range(len(l)):assert nth_element(l, 0, len(l)-1, i) == itest_nth_element()