Python利用treap實現(xiàn)雙索引的方法
:
在很多應(yīng)用場景下,我們不但需要堆的特性,例如快速知道數(shù)據(jù)最大值或最小值,同時還需要知道元素的排序信息,因此本節(jié)我們看看如何實現(xiàn)魚和熊掌如何兼得。假設(shè)我們有一系列數(shù)據(jù),它的元素由兩部分組成,一部分對應(yīng)商品的名稱,其類型為字符串,一部分對應(yīng)商品的貨存數(shù)量,類型為整形,我們既需要將商品根據(jù)其名稱排序,同時我們又需要快速查詢當(dāng)前貨存最小的商品,我們?nèi)绾卧O(shè)計相應(yīng)的算法和數(shù)據(jù)結(jié)構(gòu)來滿足這樣特性呢。
舉個例子,如下圖:
從上圖看,它對應(yīng)元素字符串是排序二叉樹,因此根節(jié)點左子樹對應(yīng)元素的字符串都小于根字符串,同時右子樹對應(yīng)的字符串都大于根節(jié)點字符串,同時每個元素還對應(yīng)著相應(yīng)商品的貨存數(shù)量,我們需要及時掌握當(dāng)前貨存最少的商品,這樣才能在其耗盡之前迅速補貨。但是從上圖可以看到,要保證字符串的排序性就得犧牲對于商品數(shù)量的小堆性質(zhì),例如上圖中water對應(yīng)的貨存與wine對應(yīng)的貨存違背了小堆的性質(zhì),現(xiàn)在問題是如何在保證字符串排序的情況下,確保數(shù)量同時能滿足小堆性質(zhì)。
首先我們先定義一下數(shù)據(jù)結(jié)構(gòu):
class Node: def __init__(self, key: str, priority: float): self._key = key self._priority = priority self._left: Node = None self._right: Node = None self._parent: Node = None @property def left(self): return self._left @property def right(self): return self._right @property def parent(self): return self._parent @left.setter def left(self, node): self._left = node if node is not None: node.parent = self @right.setter def right(self, node): self._right = node if node is not None: node.parent = self @parent.setter def parent(self, node): self._parent = node def is_root(self) -> bool: if self.parent is None: return True return False def __repr__(self): return "({}, {})".format(self._key, self._priority) def __str__(self): repr_str: str = "" repr_str += repr(self) if self.parent is not None: repr_str += " parent: " + repr(self.parent) else: repr_str += " parent: None" if self.left is not None: repr_str += " left: " + repr(self.left) else: repr_str += " left: None" if self.right is not None: repr_str += " right: " + repr(self.right) else: repr_str += " right: None" return repr_str class Treap: def __init__(self): self.root : Node = None
當(dāng)前問題是,當(dāng)上圖所示的矛盾出現(xiàn)時,我們?nèi)绾握{(diào)整,使得字符串依然保持排序性質(zhì),同時貨存數(shù)值能滿足小堆性質(zhì)。我們需要根據(jù)幾種情況采取不同操作,首先看第一種,如下圖:
從上圖看到,一種情況是父節(jié)點與左孩子在數(shù)值上違背了堆的性質(zhì),此時我們執(zhí)行一種叫右旋轉(zhuǎn)操作,
其步驟是:
Beer
節(jié)點逆時針旋轉(zhuǎn),替換其父節(jié)點;- 父節(jié)點
Cabbage
順時針旋轉(zhuǎn),成為Beer
的右孩子節(jié)點; - 原來
Beer
的右孩子節(jié)點轉(zhuǎn)變?yōu)?code>Cabbage的左孩子節(jié)點;
完成后結(jié)果如下圖所示:
可以看到,此時字符串依然保持排序二叉樹性質(zhì),同時數(shù)值對應(yīng)的小堆性質(zhì)也得到了滿足。
我們看看代碼實現(xiàn):
class Treap: def __init__(self): self._root: Node = None def right_rotate(self, x: Node): if x is None or x.is_root() is True: return y = x.parent if y.left != x: # 必須是左孩子才能右旋轉(zhuǎn) return p = y.parent if p is not None: # 執(zhí)行右旋轉(zhuǎn) if p.left == y: p.left = x else: p.right = x else: self._root = x y.left = x.right x.right = y
接下來我們構(gòu)造一些數(shù)據(jù)測試一下上面的實現(xiàn)是否正確:
def setup_right_rotate(): flour: Node = Node("Flour", 10) cabbage: Node = Node("Cabbage", 77) beer: Node = Node("Beer", 76) bacon: Node = Node("Bacon", 95) butter: Node = Node("Butter", 86) flour.parent = None flour.left = cabbage flour.right = None cabbage.left = beer beer.left = bacon beer.right = butter return flour, beer def print_treap(n: Node): if n is None: return print(n) print_treap(n.left) print_treap(n.right) treap = Treap() root, x , cabbage = setup_right_rotate() print("---------before right rotate---------:") print_treap(root) treap.right_rotate(x) print("-------after right rotate-------") print_treap(root)
上面代碼執(zhí)行后輸出內(nèi)容如下:
---------before right rotate---------:
(Flour, 10) parent: None left: (Cabbage, 77) right: None
(Cabbage, 77) parent: (Flour, 10) left: (Beer, 76) right: (Eggs, 129)
(Beer, 76) parent: (Cabbage, 77) left: (Bacon, 95) right: (Butter, 86)
(Bacon, 95) parent: (Beer, 76) left: None right: None
(Butter, 86) parent: (Beer, 76) left: None right: None
(Eggs, 129) parent: (Cabbage, 77) left: None right: None
-------after right rotate-------
(Flour, 10) parent: None left: (Beer, 76) right: None
(Beer, 76) parent: (Flour, 10) left: (Bacon, 95) right: (Cabbage, 77)
(Bacon, 95) parent: (Beer, 76) left: None right: None
(Cabbage, 77) parent: (Beer, 76) left: (Butter, 86) right: (Eggs, 129)
(Butter, 86) parent: (Cabbage, 77) left: None right: None
(Eggs, 129) parent: (Cabbage, 77) left: None right: None
對比右旋轉(zhuǎn)前后輸出的二叉樹看,旋轉(zhuǎn)后的二叉樹打印信息的確跟上面我們旋轉(zhuǎn)后對應(yīng)的圖像是一致的。接下來我們實現(xiàn)左旋轉(zhuǎn),先把上圖中cabbage
節(jié)點對應(yīng)的值改成75,這樣它與父節(jié)點就違背了小堆性質(zhì):
我們要做的是:
- 把
cabbage
節(jié)點向“左”旋轉(zhuǎn)到beer
的位置; beer
的父節(jié)點設(shè)置為cabbage
;beer
的右孩子設(shè)置為cabbage
的左孩子;cabbage
的左孩子變成beer
;左旋轉(zhuǎn)后二叉樹
成形如下:
從上圖看,左旋轉(zhuǎn)后,字符串依然保持二叉樹排序性,同時數(shù)值的排放也遵守小堆原則,我們看相應(yīng)的代碼實現(xiàn):
class Treap: ... def left_rotate(self, x : Node): if x is None or x.is_root() is True: return y = x.parent if y.right is not x: # 只有右孩子才能左旋轉(zhuǎn) return p = y.parent if p is not None: if p.left is y: p.left = x else: p.right = x else: self._root = x y.right = x.left x.left = y
為了測試上面代碼實現(xiàn),我們首先把cabbage
的值修改,然后調(diào)用上面代碼:
cabbage._priority = 75 print("-------before left rotate--------") print_treap(root) treap.left_rotate(cabbage) print("-------after left rotate---------") print_treap(root)
代碼運行后輸出結(jié)果為:
-------before left rotate--------
(Flour, 10) parent: None left: (Beer, 76) right: None
(Beer, 76) parent: (Flour, 10) left: (Bacon, 95) right: (Cabbage, 75)
(Bacon, 95) parent: (Beer, 76) left: None right: None
(Cabbage, 75) parent: (Beer, 76) left: (Butter, 86) right: (Eggs, 129)
(Butter, 86) parent: (Cabbage, 75) left: None right: None
(Eggs, 129) parent: (Cabbage, 75) left: None right: None
-------after left rotate---------
(Flour, 10) parent: None left: (Cabbage, 75) right: None
(Cabbage, 75) parent: (Flour, 10) left: (Beer, 76) right: (Eggs, 129)
(Beer, 76) parent: (Cabbage, 75) left: (Bacon, 95) right: (Butter, 86)
(Bacon, 95) parent: (Beer, 76) left: None right: None
(Butter, 86) parent: (Beer, 76) left: None right: None
(Eggs, 129) parent: (Cabbage, 75) left: None right: None
輸出結(jié)果的描述與上圖左旋轉(zhuǎn)后的結(jié)果是一致的。由于Treap
相對于元素的key是排序二叉樹,因此在給定一個字符串后,我們很容易查詢字符串是否在Treap
中,其本質(zhì)就是排序二叉樹的搜索,其實現(xiàn)我們暫時忽略。
雖然查詢很簡單,但是插入節(jié)點則稍微麻煩,因為插入后,新節(jié)點與其父節(jié)點可能會違背小堆性質(zhì),因此在完成插入后,我們還需使用上面實現(xiàn)的左旋轉(zhuǎn)或右旋轉(zhuǎn)來進行調(diào)整。
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