What is the best pattern to reverse a string in-place?

Using two-pointer scanning with invariant tracking is optimal for reversing a string array in-place with minimal memory.

Can I use extra memory to simplify reversing a string?

No, the problem explicitly requires O(1) extra memory, so all swaps must be performed in-place.

How do I handle a string array of length 1 or 0?

These arrays are already reversed by definition, so no swaps are needed but pointer logic should still handle them safely.

What is the time complexity of this two-pointer approach?

The approach runs in O(n) time since each character is visited exactly once during swaps.

Why is two-pointer scanning preferred for the Reverse String problem?

It guarantees in-place reversal with O(1) extra memory and avoids common pitfalls like off-by-one errors or unnecessary array copying.

#344

Easy

auto_awesome双·指针·invariant

LeetCode 题解工作台

反转字符串

编写一个函数，其作用是将输入的字符串反转过来。输入字符串以字符数组 s 的形式给出。不要给另外的数组分配额外的空间，你必须原地修改输入数组、使用 O(1) 的额外空间解决这一问题。示例 1：输入： s = ["h","e","l","l","o"] 输出： ["o","l","l","e…

双指针字符串

题目描述

编写一个函数，其作用是将输入的字符串反转过来。输入字符串以字符数组 s 的形式给出。

不要给另外的数组分配额外的空间，你必须原地修改输入数组、使用 O(1) 的额外空间解决这一问题。

示例 1：

输入：s = ["h","e","l","l","o"]
输出：["o","l","l","e","h"]

示例 2：

输入：s = ["H","a","n","n","a","h"]
输出：["h","a","n","n","a","H"]

提示：

1 <= s.length <= 10⁵
s[i] 都是 ASCII 码表中的可打印字符

lightbulb

解题思路

方法一：双指针

我们用两个指针 $i$ 和 $j$ ，初始时分别指向数组的首尾，每次将 $i$ 和 $j$ 对应的元素交换，然后 $i$ 向后移动， $j$ 向前移动，直到 $i$ 和 $j$ 相遇。

时间复杂度 $O(n)$ ，其中 $n$ 是数组的长度。空间复杂度 $O(1)$ 。

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class Solution:
    def reverseString(self, s: List[str]) -> None:
        i, j = 0, len(s) - 1
        while i < j:
            s[i], s[j] = s[j], s[i]
            i, j = i + 1, j - 1

speed

复杂度分析

指标	值
时间	complexity is O(n) because each character is visited once in a single pass. Space complexity is O(1) since all swaps are in-place and no additional arrays are created.
空间	Depends on the final approach

psychology

面试官常问的追问

外企场景

question_mark
Looks for correct two-pointer initialization and proper index tracking.
question_mark
Checks that swaps are performed in-place without auxiliary storage.
question_mark
Tests edge cases like empty arrays or arrays with one character.

warning

常见陷阱

外企场景

error
Forgetting to move both pointers after a swap, causing infinite loops.
error
Using extra arrays instead of in-place modification.
error
Mismanaging indices, leading to off-by-one errors or partial reversal.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Reverse a linked list using two-pointer or recursive swapping.
arrow_right_alt
Reverse only a substring or a section of the array in-place.
arrow_right_alt
Reverse words in a sentence while maintaining character order within words.

help

常见问题

外企场景

继续练习

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