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feat: Traditional Chinese version (#1163)
* First commit * Update mkdocs.yml * Translate all the docs to traditional Chinese * Translate the code files. * Translate the docker file * Fix mkdocs.yml * Translate all the figures from SC to TC * 二叉搜尋樹 -> 二元搜尋樹 * Update terminology. * Update terminology * 构造函数/构造方法 -> 建構子 异或 -> 互斥或 * 擴充套件 -> 擴展 * constant - 常量 - 常數 * 類 -> 類別 * AVL -> AVL 樹 * 數組 -> 陣列 * 係統 -> 系統 斐波那契數列 -> 費波那契數列 運算元量 -> 運算量 引數 -> 參數 * 聯絡 -> 關聯 * 麵試 -> 面試 * 面向物件 -> 物件導向 歸併排序 -> 合併排序 范式 -> 範式 * Fix 算法 -> 演算法 * 錶示 -> 表示 反碼 -> 一補數 補碼 -> 二補數 列列尾部 -> 佇列尾部 區域性性 -> 區域性 一摞 -> 一疊 * Synchronize with main branch * 賬號 -> 帳號 推匯 -> 推導 * Sync with main branch * First commit * Update mkdocs.yml * Translate all the docs to traditional Chinese * Translate the code files. * Translate the docker file * Fix mkdocs.yml * Translate all the figures from SC to TC * 二叉搜尋樹 -> 二元搜尋樹 * Update terminology * 构造函数/构造方法 -> 建構子 异或 -> 互斥或 * 擴充套件 -> 擴展 * constant - 常量 - 常數 * 類 -> 類別 * AVL -> AVL 樹 * 數組 -> 陣列 * 係統 -> 系統 斐波那契數列 -> 費波那契數列 運算元量 -> 運算量 引數 -> 參數 * 聯絡 -> 關聯 * 麵試 -> 面試 * 面向物件 -> 物件導向 歸併排序 -> 合併排序 范式 -> 範式 * Fix 算法 -> 演算法 * 錶示 -> 表示 反碼 -> 一補數 補碼 -> 二補數 列列尾部 -> 佇列尾部 區域性性 -> 區域性 一摞 -> 一疊 * Synchronize with main branch * 賬號 -> 帳號 推匯 -> 推導 * Sync with main branch * Update terminology.md * 操作数量(num. of operations)-> 操作數量 * 字首和->前綴和 * Update figures * 歸 -> 迴 記憶體洩漏 -> 記憶體流失 * Fix the bug of the file filter * 支援 -> 支持 Add zh-Hant/README.md * Add the zh-Hant chapter covers. Bug fixes. * 外掛 -> 擴充功能 * Add the landing page for zh-Hant version * Unify the font of the chapter covers for the zh, en, and zh-Hant version * Move zh-Hant/ to zh-hant/ * Translate terminology.md to traditional Chinese
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71
zh-hant/codes/python/chapter_heap/heap.py
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71
zh-hant/codes/python/chapter_heap/heap.py
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"""
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File: heap.py
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Created Time: 2023-02-23
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Author: krahets (krahets@163.com)
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"""
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import sys
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from pathlib import Path
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sys.path.append(str(Path(__file__).parent.parent))
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from modules import print_heap
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import heapq
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def test_push(heap: list, val: int, flag: int = 1):
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heapq.heappush(heap, flag * val) # 元素入堆積
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print(f"\n元素 {val} 入堆積後")
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print_heap([flag * val for val in heap])
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def test_pop(heap: list, flag: int = 1):
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val = flag * heapq.heappop(heap) # 堆積頂元素出堆積
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print(f"\n堆積頂元素 {val} 出堆積後")
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print_heap([flag * val for val in heap])
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"""Driver Code"""
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if __name__ == "__main__":
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# 初始化小頂堆積
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min_heap, flag = [], 1
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# 初始化大頂堆積
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max_heap, flag = [], -1
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print("\n以下測試樣例為大頂堆積")
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# Python 的 heapq 模組預設實現小頂堆積
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# 考慮將“元素取負”後再入堆積,這樣就可以將大小關係顛倒,從而實現大頂堆積
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# 在本示例中,flag = 1 時對應小頂堆積,flag = -1 時對應大頂堆積
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# 元素入堆積
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test_push(max_heap, 1, flag)
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test_push(max_heap, 3, flag)
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test_push(max_heap, 2, flag)
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test_push(max_heap, 5, flag)
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test_push(max_heap, 4, flag)
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# 獲取堆積頂元素
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peek: int = flag * max_heap[0]
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print(f"\n堆積頂元素為 {peek}")
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# 堆積頂元素出堆積
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test_pop(max_heap, flag)
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test_pop(max_heap, flag)
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test_pop(max_heap, flag)
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test_pop(max_heap, flag)
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test_pop(max_heap, flag)
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# 獲取堆積大小
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size: int = len(max_heap)
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print(f"\n堆積元素數量為 {size}")
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# 判斷堆積是否為空
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is_empty: bool = not max_heap
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print(f"\n堆積是否為空 {is_empty}")
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# 輸入串列並建堆積
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# 時間複雜度為 O(n) ,而非 O(nlogn)
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min_heap = [1, 3, 2, 5, 4]
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heapq.heapify(min_heap)
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print("\n輸入串列並建立小頂堆積後")
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print_heap(min_heap)
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137
zh-hant/codes/python/chapter_heap/my_heap.py
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137
zh-hant/codes/python/chapter_heap/my_heap.py
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"""
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File: my_heap.py
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Created Time: 2023-02-23
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Author: krahets (krahets@163.com)
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"""
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import sys
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from pathlib import Path
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sys.path.append(str(Path(__file__).parent.parent))
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from modules import print_heap
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class MaxHeap:
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"""大頂堆積"""
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def __init__(self, nums: list[int]):
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"""建構子,根據輸入串列建堆積"""
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# 將串列元素原封不動新增進堆積
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self.max_heap = nums
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# 堆積化除葉節點以外的其他所有節點
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for i in range(self.parent(self.size() - 1), -1, -1):
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self.sift_down(i)
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def left(self, i: int) -> int:
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"""獲取左子節點的索引"""
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return 2 * i + 1
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def right(self, i: int) -> int:
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"""獲取右子節點的索引"""
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return 2 * i + 2
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def parent(self, i: int) -> int:
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"""獲取父節點的索引"""
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return (i - 1) // 2 # 向下整除
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def swap(self, i: int, j: int):
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"""交換元素"""
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self.max_heap[i], self.max_heap[j] = self.max_heap[j], self.max_heap[i]
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def size(self) -> int:
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"""獲取堆積大小"""
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return len(self.max_heap)
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def is_empty(self) -> bool:
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"""判斷堆積是否為空"""
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return self.size() == 0
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def peek(self) -> int:
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"""訪問堆積頂元素"""
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return self.max_heap[0]
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def push(self, val: int):
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"""元素入堆積"""
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# 新增節點
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self.max_heap.append(val)
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# 從底至頂堆積化
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self.sift_up(self.size() - 1)
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def sift_up(self, i: int):
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"""從節點 i 開始,從底至頂堆積化"""
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while True:
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# 獲取節點 i 的父節點
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p = self.parent(i)
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# 當“越過根節點”或“節點無須修復”時,結束堆積化
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if p < 0 or self.max_heap[i] <= self.max_heap[p]:
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break
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# 交換兩節點
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self.swap(i, p)
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# 迴圈向上堆積化
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i = p
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def pop(self) -> int:
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"""元素出堆積"""
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# 判空處理
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if self.is_empty():
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raise IndexError("堆積為空")
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# 交換根節點與最右葉節點(交換首元素與尾元素)
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self.swap(0, self.size() - 1)
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# 刪除節點
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val = self.max_heap.pop()
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# 從頂至底堆積化
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self.sift_down(0)
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# 返回堆積頂元素
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return val
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def sift_down(self, i: int):
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"""從節點 i 開始,從頂至底堆積化"""
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while True:
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# 判斷節點 i, l, r 中值最大的節點,記為 ma
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l, r, ma = self.left(i), self.right(i), i
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if l < self.size() and self.max_heap[l] > self.max_heap[ma]:
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ma = l
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if r < self.size() and self.max_heap[r] > self.max_heap[ma]:
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ma = r
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# 若節點 i 最大或索引 l, r 越界,則無須繼續堆積化,跳出
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if ma == i:
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break
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# 交換兩節點
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self.swap(i, ma)
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# 迴圈向下堆積化
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i = ma
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def print(self):
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"""列印堆積(二元樹)"""
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print_heap(self.max_heap)
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"""Driver Code"""
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if __name__ == "__main__":
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# 初始化大頂堆積
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max_heap = MaxHeap([9, 8, 6, 6, 7, 5, 2, 1, 4, 3, 6, 2])
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print("\n輸入串列並建堆積後")
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max_heap.print()
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# 獲取堆積頂元素
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peek = max_heap.peek()
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print(f"\n堆積頂元素為 {peek}")
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# 元素入堆積
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val = 7
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max_heap.push(val)
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print(f"\n元素 {val} 入堆積後")
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max_heap.print()
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# 堆積頂元素出堆積
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peek = max_heap.pop()
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print(f"\n堆積頂元素 {peek} 出堆積後")
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max_heap.print()
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# 獲取堆積大小
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size = max_heap.size()
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print(f"\n堆積元素數量為 {size}")
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# 判斷堆積是否為空
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is_empty = max_heap.is_empty()
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print(f"\n堆積是否為空 {is_empty}")
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39
zh-hant/codes/python/chapter_heap/top_k.py
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39
zh-hant/codes/python/chapter_heap/top_k.py
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"""
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File: top_k.py
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Created Time: 2023-06-10
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Author: krahets (krahets@163.com)
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"""
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import sys
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from pathlib import Path
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sys.path.append(str(Path(__file__).parent.parent))
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from modules import print_heap
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import heapq
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def top_k_heap(nums: list[int], k: int) -> list[int]:
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"""基於堆積查詢陣列中最大的 k 個元素"""
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# 初始化小頂堆積
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heap = []
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# 將陣列的前 k 個元素入堆積
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for i in range(k):
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heapq.heappush(heap, nums[i])
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# 從第 k+1 個元素開始,保持堆積的長度為 k
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for i in range(k, len(nums)):
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# 若當前元素大於堆積頂元素,則將堆積頂元素出堆積、當前元素入堆積
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if nums[i] > heap[0]:
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heapq.heappop(heap)
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heapq.heappush(heap, nums[i])
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return heap
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"""Driver Code"""
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if __name__ == "__main__":
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nums = [1, 7, 6, 3, 2]
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k = 3
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res = top_k_heap(nums, k)
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print(f"最大的 {k} 個元素為")
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print_heap(res)
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