|
|
|
|
|
"""
|
|
|
|
|
|
File: heap.py
|
|
|
|
|
|
Created Time: 2023-02-23
|
|
|
|
|
|
Author: krahets (krahets@163.com)
|
|
|
|
|
|
"""
|
|
|
|
|
|
|
|
|
|
|
|
import sys
|
|
|
|
|
|
from pathlib import Path
|
|
|
|
|
|
|
|
|
|
|
|
sys.path.append(str(Path(__file__).parent.parent))
|
|
|
|
|
|
from modules import print_heap
|
|
|
|
|
|
|
|
|
|
|
|
import heapq
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
def test_push(heap: list, val: int, flag: int = 1):
|
|
|
|
|
|
heapq.heappush(heap, flag * val) # 元素入堆積
|
|
|
|
|
|
print(f"\n元素 {val} 入堆積後")
|
|
|
|
|
|
print_heap([flag * val for val in heap])
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
def test_pop(heap: list, flag: int = 1):
|
|
|
|
|
|
val = flag * heapq.heappop(heap) # 堆積頂元素出堆積
|
|
|
|
|
|
print(f"\n堆積頂元素 {val} 出堆積後")
|
|
|
|
|
|
print_heap([flag * val for val in heap])
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
"""Driver Code"""
|
|
|
|
|
|
if __name__ == "__main__":
|
|
|
|
|
|
# 初始化小頂堆積
|
|
|
|
|
|
min_heap, flag = [], 1
|
|
|
|
|
|
# 初始化大頂堆積
|
|
|
|
|
|
max_heap, flag = [], -1
|
|
|
|
|
|
|
|
|
|
|
|
print("\n以下測試樣例為大頂堆積")
|
|
|
|
|
|
# Python 的 heapq 模組預設實現小頂堆積
|
|
|
|
|
|
# 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
|
|
|
|
|
|
# 在本示例中，flag = 1 時對應小頂堆積，flag = -1 時對應大頂堆積
|
|
|
|
|
|
|
|
|
|
|
|
# 元素入堆積
|
|
|
|
|
|
test_push(max_heap, 1, flag)
|
|
|
|
|
|
test_push(max_heap, 3, flag)
|
|
|
|
|
|
test_push(max_heap, 2, flag)
|
|
|
|
|
|
test_push(max_heap, 5, flag)
|
|
|
|
|
|
test_push(max_heap, 4, flag)
|
|
|
|
|
|
|
|
|
|
|
|
# 獲取堆積頂元素
|
|
|
|
|
|
peek: int = flag * max_heap[0]
|
|
|
|
|
|
print(f"\n堆積頂元素為 {peek}")
|
|
|
|
|
|
|
|
|
|
|
|
# 堆積頂元素出堆積
|
|
|
|
|
|
test_pop(max_heap, flag)
|
|
|
|
|
|
test_pop(max_heap, flag)
|
|
|
|
|
|
test_pop(max_heap, flag)
|
|
|
|
|
|
test_pop(max_heap, flag)
|
|
|
|
|
|
test_pop(max_heap, flag)
|
|
|
|
|
|
|
|
|
|
|
|
# 獲取堆積大小
|
|
|
|
|
|
size: int = len(max_heap)
|
|
|
|
|
|
print(f"\n堆積元素數量為 {size}")
|
|
|
|
|
|
|
|
|
|
|
|
# 判斷堆積是否為空
|
|
|
|
|
|
is_empty: bool = not max_heap
|
|
|
|
|
|
print(f"\n堆積是否為空 {is_empty}")
|
|
|
|
|
|
|
|
|
|
|
|
# 輸入串列並建堆積
|
|
|
|
|
|
# 時間複雜度為 O(n) ，而非 O(nlogn)
|
|
|
|
|
|
min_heap = [1, 3, 2, 5, 4]
|
|
|
|
|
|
heapq.heapify(min_heap)
|
|
|
|
|
|
print("\n輸入串列並建立小頂堆積後")
|
|
|
|
|
|
print_heap(min_heap)
|
