算法阅读笔记

Best Time to Buy/Sell Stocks

For the one-transaction stock problem, scan once while tracking the minimum price seen so far and the best profit achievable by selling today.

类型算法分类Classic Problems深度初级典型复杂度Varies by implementation / input constraints

简短面试回答

先从这个心智模型开始

For the one-transaction stock problem, scan once while tracking the minimum price seen so far and the best profit achievable by selling today.

简单解释

它如何工作，不堆砌不必要的术语

The buy must occur before the sell, so only past minimums are valid buy candidates.

One pass gives O(n) time and O(1) extra space.

执行过程

把算法读成一连串决策

Restate constraints and edge cases.

Identify the pattern and invariant.

Walk through a small example before coding.

记住一个小例子

Read a tiny input by hand and write down the state after every important operation. For Best Time to Buy/Sell Stocks, the goal is to make the invariant visible rather than to memorize code.

易读的 JavaScript 实现JavaScript

function maxProfit(prices) {
  let minPrice = Infinity;
  let best = 0;
  for (const price of prices) {
    minPrice = Math.min(minPrice, price);
    best = Math.max(best, price - minPrice);
  }
  return best;
}

如何阅读这段代码

不要死记语法。跟踪每一行中发生变化的状态，并把它与上面描述的不变量联系起来。

复杂度

Varies by implementation / input constraints

Do not quote this complexity mechanically. Ask which operation or input dimension it describes, then derive the cost from the loops, recursion depth, data-structure operations and any preprocessing assumptions.

何时使用

适用场景

Use the recognized pattern when its invariant matches the problem constraints.

常见错误

通常哪里会出错

Jumping to code before defining the invariant.
Optimizing before producing a correct baseline.

面试中如何解释

专业回答结构

Do not sort prices; sorting destroys time order. Track the best historical buy price instead.

问题→核心思路→不变量 / 状态→Big-O→权衡

快速面试检查

What problem does Best Time to Buy/Sell Stocks solve?

For the one-transaction stock problem, scan once while tracking the minimum price seen so far and the best profit achievable by selling today.

关于复杂度应该提到什么？

Its typical cost here is Varies by implementation / input constraints, but a complete answer should connect that cost to the representation, input constraints and the dominant operations.

最常见的错误是什么？

Jumping to code before defining the invariant.

继续阅读Problem patternsEdge casesComplexity trade-offs

function maxProfit(prices) { let minPrice = Infinity; let best = 0; for (const price of prices) { minPrice = Math.min(minPrice, price); best = Math.max(best, price - minPrice); } return best; }