科技公司 SWE 在线评估：双指针与滑动窗口从头讲清
SWE Online Assessment: Two Pointers and Sliding Window, From Scratch
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摘要 Summary
双指针和滑动窗口，能把很多 O(n^2) 的暴力想法压成一次 O(n) 扫描。这篇文章讲你真正会遇到的三种形态——两端相向、快慢指针、可伸缩窗口，并用一个大家最容易卡住的可变长度窗口例子把它写清楚。
Two pointers and sliding window turn many O(n^2) brute-force ideas into a single O(n) pass. This guide covers the three shapes you actually see — opposite ends, fast and slow, and a growing/shrinking window — with a clean example of the variable-size window that trips people up most.
双指针和滑动窗口本质上是同一个想法：不去枚举每一对元素或每一个子数组，而是让一两个下标在数据上走一遍，边走边只维护刚好够用的状态。就是这一点改变，把 O(n^2) 的暴力变成了 O(n) 的一遍扫描，也是为什么这类题大约每六道就会出现一道。
这篇文章整理自公开、广为人知的算法题型，而不是某家公司的真实原题。把它当成一张练习地图：先记住每种题的识别信号，每个模式手里留一份干净的参考解法，剩下的时间用来把题读懂，读到能一眼选对套路。
你真正会遇到的三种形态| The Three Shapes You Actually See
两端相向：一个指针从左端、一个从右端，往中间靠拢——常见于有序数组里的两数之和、回文判断。
快慢指针：两个指针以不同速度走同一个序列——用来判环、找中点。
滑动窗口：左右两个边界圈出一个窗口，右边界扩张纳入新元素，约束被破坏时左边界收缩。
固定窗口与可变窗口| Fixed Versus Variable Windows
固定窗口的长度 k 是已知的：右边加入新元素，左边移出滑出去的那个，每一步读一次答案。
可变窗口先一直扩张，直到违反某个条件，再从左边收缩，直到条件重新成立。
大多数人卡住的正是可变窗口，因为「什么时候收缩」和「什么时候记录答案」这两步都必须踩得很准。
一个例子：无重复字符的最长子串| A Worked Example: Longest Substring Without Repeats
这是可变窗口的经典题。右边界每次向右吃进一个字符；一旦这个新字符已经在窗口里，就从左边收缩，直到它不再重复。每一步结束时窗口总是合法的，所以可以放心地记录它的长度。
def longest_unique_substring(s):
seen = {} # char -> most recent index
left = 0
best = 0
for right, ch in enumerate(s):
# if ch is already in the window, jump left past its last position
if ch in seen and seen[ch] >= left:
left = seen[ch] + 1
seen[ch] = right
best = max(best, right - left + 1)
return best注意 seen[ch] >= left 这个判断。少了它，一个来自当前窗口之前的旧下标会把左指针往回拉，把答案弄坏。正是这种小条件，面试官盯得最紧，所以动手前先把不变量说出来——「left 到 right 之间的字符互不相同」。
最容易失分的几个点| Common Mistakes That Cost Marks
把左指针往回移，或者忘了旧下标可能已经落在当前窗口之外。
在错误的时刻记录答案——在收缩之前就记，而不是等窗口重新合法之后。
出于习惯写了两层嵌套循环，其实一次窗口扫描就是 O(n)。
做两端相向的题时，忘了两数之和的逻辑成立的前提是数组已经有序。
一个够用的练习计划| A Focused Practice Plan
把一道固定窗口题和一道可变窗口题连着做，真正体会两者的区别。
每道窗口题，都在循环前用注释写下不变量。
练习用嘴说清楚：窗口什么时候扩张、什么时候收缩。
把快慢指针判环的思路从头自己推一遍，别只当成一个背下来的技巧。
常见问题 FAQ
什么时候双指针比哈希表更合适？
When is two pointers better than a hash map?
当输入已经有序、或者排序代价很小时，双指针常常能以 O(1) 额外空间解决配对和区间问题；而哈希表是用内存换「不需要有序」。
怎么判断窗口该用固定长度还是可变长度？
How do I know whether a window should be fixed or variable?
如果题目直接给了长度 k，就是固定窗口。如果问的是满足某个条件的最长或最短窗口，就是可变窗口，用先扩张后收缩的写法。
为什么我的滑动窗口在边界用例上会算错？
Why does my sliding window give the wrong answer on edge cases?
通常是收缩条件或记录答案的时机差了一。回头确认：在你读取窗口大小的那一刻，窗口是不是合法的。
How This Article Was Produced
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