chore: update 2026-04-21

solutions/0054.PokerHands/euler_54_better.py

@@ -0,0 +1,101 @@
+import time
+from collections import Counter
+from pathlib import Path
+
+
+def timer(func):
+    def wrapper(*args, **kwargs):
+        start_time = time.time()
+        result = func(*args, **kwargs)
+        end_time = time.time()
+        elapsed_time = end_time - start_time
+        print(f"{func.__name__} time: {elapsed_time:.6f} seconds")
+        return result
+
+    return wrapper
+
+
+# 牌面数值映射
+VAL = {
+    "2": 2,
+    "3": 3,
+    "4": 4,
+    "5": 5,
+    "6": 6,
+    "7": 7,
+    "8": 8,
+    "9": 9,
+    "T": 10,
+    "J": 11,
+    "Q": 12,
+    "K": 13,
+    "A": 14,
+}
+
+
+def hand_key(cards: tuple[str, ...]) -> tuple[int, ...]:
+    """
+    将一手5张牌映射为规范元组。
+    元组的字典序即牌力大小，可直接用 > / < 比较。
+    """
+    # 1. 提取数值并降序排列
+    nums = sorted([VAL[c[0]] for c in cards], reverse=True)
+    suits = {c[1] for c in cards}
+
+    # 2. 判断同花与顺子
+    is_flush = len(suits) == 1
+    is_straight = len(set(nums)) == 5 and nums[0] - nums[4] == 4
+
+    # 3. 处理 A-2-3-4-5 顺子（A作为1，5-high）
+    if nums == [14, 5, 4, 3, 2]:
+        is_straight = True
+        nums = [5, 4, 3, 2, 1]
+
+    # 4. 统计出现次数，并按 (次数, 牌面) 降序排列
+    #    这是整个方法的数学核心：多重集的字典序
+    counts = Counter(nums)
+    by_count = sorted(counts.items(), key=lambda x: (x[1], x[0]), reverse=True)
+
+    # 5. 按牌型返回规范元组
+    if is_flush and is_straight:
+        return (9, nums[0])  # 同花顺（皇家同花顺只是顶牌为A）
+    if by_count[0][1] == 4:
+        return (8, by_count[0][0], by_count[1][0])  # 四条
+    if by_count[0][1] == 3 and len(by_count) == 2:
+        return (7, by_count[0][0], by_count[1][0])  # 葫芦
+    if is_flush:
+        return (6, *nums)  # 同花：比全部5张降序
+    if is_straight:
+        return (5, nums[0])  # 顺子：比顶牌
+    if by_count[0][1] == 3:
+        return (4, by_count[0][0], by_count[1][0], by_count[2][0])  # 三条
+    if by_count[0][1] == 2 and len(by_count) == 3:
+        return (3, by_count[0][0], by_count[1][0], by_count[2][0])  # 两对
+    if by_count[0][1] == 2:
+        return (
+            2,
+            by_count[0][0],
+            by_count[1][0],
+            by_count[2][0],
+            by_count[3][0],
+        )  # 一对
+    return (1, *nums)  # 高牌：比全部5张降序
+
+
+def read_hands(path: Path):
+    with open(path) as f:
+        for line in f:
+            cards = line.strip().split()
+            if len(cards) == 10:
+                yield tuple(cards[:5]), tuple(cards[5:])
+
+
+@timer
+def main() -> int:
+    data_path = Path(__file__).parent / "0054_poker.txt"
+    wins = sum(1 for p1, p2 in read_hands(data_path) if hand_key(p1) > hand_key(p2))
+    return wins
+
+
+if __name__ == "__main__":
+    print(main())