面试题库

14 天用 Agent 搭一条 AI 短剧生产线

D6 剪辑台：用 ffmpeg 把素材合成一集竖屏成片

• 把多个视频片段拼成一条完整的视频，什么时候可以不重新编码，什么时候必须重编码？When can you concatenate video segments without re-encoding, and when must you re-encode?

  国内高频海外高频基础#ffmpeg#encoding#media-pipeline

  分析过程 · 先想清楚再作答

  1. 这是一道概念送分题，但送分的是「说出前提」的人。答「用 concat 就行」和答「参数一致才能流拷贝」，在面试官眼里是两个水平。
  2. 判据只有一条：拼接是不是只需要把数据包按顺序搬进新容器。只搬不算，就能流拷贝；只要有任何一帧画面是新算出来的，就必须重编码。
  3. 流拷贝的前提要能背出来：分辨率、帧率、像素格式、编码器、音频采样率、声道数全部一致。差一项，产物要么花屏掉音，要么时长错乱。
  4. 必须重编码的典型场景：转场（那几帧是新画面）、缩放补边到统一画布、混入新的音轨、改变编码参数。AI 生成的素材尺寸和音轨天然不一致，所以实际工程里几乎总要先归一化。
  5. 结论落在一个组合拳上：每一镜单独走一次滤镜图做归一化，然后用流拷贝把参数已经一致的片段拼起来。重编码的总量还是一遍，但换来了对每一镜的完全控制。
  6. 可预期的追问是「怎么判断素材参数一不一致」。答：用 ffprobe 把每段的关键字段读出来做一次比对，不一致就走归一化，这一步也是自动化流水线里必须有的前置检查。

  How to reason about it · think before answering

  1. This is a giveaway concept question, but it only gives points to people who state the precondition. 'Just use concat' and 'stream copy requires identical parameters' read as two different levels.
  2. There is exactly one criterion: does concatenation only need to move packets into a new container in order? If yes, stream copy works. If even one frame has to be newly computed, you must re-encode.
  3. Be able to recite the preconditions: resolution, frame rate, pixel format, codec, audio sample rate and channel layout must all match. Miss one and you get corruption, dropped audio, or a broken duration.
  4. Cases that force re-encoding: transitions (those frames are new), scaling and padding to a common canvas, mixing in a new audio track, or changing encoding parameters. AI-generated material varies in size and often lacks audio, so normalization is almost always required in practice.
  5. The conclusion is a combination: normalize each shot with its own filter graph pass, then stream-copy the now-identical segments together. Total re-encoding is still one pass, but you gain full control over each shot.
  6. Expect the follow-up 'how do you know whether the parameters match'. Answer: read the key fields of each segment with ffprobe and compare. That precheck belongs in any automated pipeline.

  答题要点

  • 判据是拼接是否只需要搬数据包：只搬就能流拷贝，有新算出来的帧就必须重编码。
  • 流拷贝的前提：分辨率、帧率、像素格式、编码器、采样率、声道数全部一致。
  • 转场、缩放补边、混入新音轨、改编码参数，这几类一定要重编码。
  • 实践中的组合拳：先逐镜归一化，再流拷贝拼接，重编码总量仍是一遍。
  • 用 ffprobe 比对各段参数，作为流水线里的前置检查。

  Key points

  • The criterion is whether concatenation only moves packets: if so, stream copy; if any frame is newly computed, re-encode.
  • Stream-copy preconditions: identical resolution, frame rate, pixel format, codec, sample rate and channel layout.
  • Transitions, scale-and-pad, mixing a new audio track, and changing encoding parameters all force re-encoding.
  • The practical combination: normalize per shot first, then stream-copy concatenate. Total re-encoding stays at one pass.
  • Use ffprobe to compare segment parameters as a pipeline precheck.

D13 发行：多平台规格适配、封面与标题生成、批量导出与数据回收

• 同一条视频要发多个平台，每个平台规格不同，你会怎么设计导出流程才能少转码？The same video has to be published to several platforms with different specs. How do you design the export flow to minimize transcoding?

  国内高频海外高频进阶#media-pipeline#ffmpeg

  分析过程 · 先想清楚再作答

  1. 这题在考你分不分得清转码与封装。答成「按每个平台各渲染一遍」的人不是不会写代码，是没意识到有损编码每转一次就掉一次画质。
  2. 先把两个词分开：转码是重新解码再编码，画面数据真的被压了一遍；封装只是把已编好的码流换个容器，一个字节都没动。前者要几秒到几十秒并且掉画质，后者几十毫秒且无损。
  3. 然后给流程：先渲染一份母版，参数取所有目标平台的交集里最保守的一档；之后每个平台走一次判定函数，能流复制就流复制。换容器、加 faststart、按时长截断都属于流复制的范围。
  4. 必须重编码的情况要能背出来：分辨率越界要缩放、编码格式不被接受、帧率超范围、文件大小超限要降码率。除此之外都不该重编码——尤其时长超限这一条最容易被误判，其实 -t 配流复制就能切。
  5. 补一条工程判据：判定函数要返回理由列表，不只是布尔值。出片之后有人问「为什么这个平台转了码」，你要能拿日志回答，而不是重新读一遍代码。
  6. 可预期的追问是「怎么证明真的少转了」。答案是打印一个编码次数计数器，并同时给出朴素做法的次数做对照——没有对照的数字说服不了任何人。

  How to reason about it · think before answering

  1. This question tests whether you distinguish transcoding from remuxing. Rendering once per platform is not a coding failure, it is a failure to notice that every lossy re-encode costs quality.
  2. Separate the two: transcoding decodes and re-encodes, so the picture data is genuinely recompressed; remuxing just moves an already-encoded bitstream into another container without touching a byte. One takes seconds and loses quality, the other takes milliseconds and is lossless.
  3. Then give the flow: render one master using the most conservative parameters that satisfy every target, then run each platform through a decision function and stream-copy whenever possible. Container changes, faststart and duration trims all stay within stream copy.
  4. Know the cases that truly require re-encoding: out-of-range resolution, an unaccepted codec, a frame rate outside the allowed band, and a file that exceeds the size cap. Duration is the one people misjudge most, since -t with stream copy already trims it.
  5. Add an engineering rule: the decision function should return a list of reasons, not just a boolean. When someone asks why a platform got re-encoded, you answer from the log rather than rereading the code.
  6. Expect the follow-up: how do you prove it? Print an encode counter alongside the count the naive approach would have produced. A number without a baseline convinces nobody.

  答题要点

  • 分清转码与封装：换容器、加 faststart、按时长截断都可以流复制
  • 一次渲染母版，参数取所有目标平台约束的最保守交集
  • 只有分辨率越界、编码不被接受、帧率超范围、体积超限才必须重编码
  • 判定函数返回理由列表，让每次重编码都能被解释
  • 打印编码次数计数器并与朴素做法做对照，才算证明少转了码

  Key points

  • Separate transcode from remux: container swaps, faststart and duration trims are all stream copies
  • Render one master using the most conservative intersection of all target constraints
  • Re-encode only for out-of-range resolution, unaccepted codec, out-of-band frame rate, or oversize files
  • Have the decision function return reasons so every re-encode can be explained
  • Print an encode counter next to the naive baseline to prove the saving