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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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