What is the core pattern in Reverse String II?

The main pattern is two-pointer scanning with invariant tracking, reversing only the first k characters in every 2k block.

How do I handle the last segment if it has fewer than k characters?

Reverse all remaining characters in that segment, ensuring no indices go out of bounds.

Is using extra space necessary for this problem?

No, you can convert the string to a character array and perform in-place swaps using two pointers to maintain O(N) space.

Can this pattern be applied to arrays of numbers?

Yes, the two-pointer invariant reversal approach works for arrays of any elements, not just strings.

Why not reverse the entire 2k block at once?

Reversing the whole block would violate the problem requirement of reversing only the first k characters, producing incorrect output.

#541

Easy

auto_awesome双·指针·invariant

LeetCode 题解工作台

反转字符串 II

给定一个字符串 s 和一个整数 k ，从字符串开头算起，每计数至 2k 个字符，就反转这 2k 字符中的前 k 个字符。如果剩余字符少于 k 个，则将剩余字符全部反转。如果剩余字符小于 2k 但大于或等于 k 个，则反转前 k 个字符，其余字符保持原样。示例 1：输入： s = "abcde…

双指针字符串

题目描述

给定一个字符串 s 和一个整数 k，从字符串开头算起，每计数至 2k 个字符，就反转这 2k 字符中的前 k 个字符。

如果剩余字符少于 k 个，则将剩余字符全部反转。
如果剩余字符小于 2k 但大于或等于 k 个，则反转前 k 个字符，其余字符保持原样。

示例 1：

输入：s = "abcdefg", k = 2
输出："bacdfeg"

示例 2：

输入：s = "abcd", k = 2
输出："bacd"

提示：

1 <= s.length <= 10⁴
s 仅由小写英文组成
1 <= k <= 10⁴

lightbulb

解题思路

方法一：双指针

我们可以遍历字符串 $\textit{s}$ ，每次遍历 $\textit{2k}$ 个字符，然后利用双指针技巧，对这 $\textit{2k}$ 个字符中的前 $\textit{k}$ 个字符进行反转。

时间复杂度 $O(n)$ ，空间复杂度 $O(n)$ 。其中 $n$ 为字符串 $\textit{s}$ 的长度。

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class Solution:
    def reverseStr(self, s: str, k: int) -> str:
        cs = list(s)
        for i in range(0, len(cs), 2 * k):
            cs[i : i + k] = reversed(cs[i : i + k])
        return "".join(cs)

speed

复杂度分析

指标	值
时间	complexity is O(N) because each character is visited at most once during the swaps. Space complexity is O(N) if using a character array for in-place swaps; otherwise, extra space is minimal beyond the original string.
空间	O(N)

psychology

面试官常问的追问

外企场景

question_mark
Look for proper handling of blocks smaller than k to test boundary understanding.
question_mark
Ensure the candidate uses two pointers rather than naive slicing to avoid extra memory usage.
question_mark
Watch if the candidate correctly iterates in 2k steps rather than per-character to maintain efficiency.

warning

常见陷阱

外企场景

error
Reversing more than k characters in a 2k block and corrupting the second half.
error
Failing to handle the last partial segment correctly when its length is less than k.
error
Using string concatenation inside loops, leading to O(N^2) time instead of O(N).

swap_horiz

进阶变体

外企场景

arrow_right_alt
Reverse every first k characters in 3k blocks instead of 2k, requiring adjusted pointer increments.
arrow_right_alt
Perform conditional reversal based on character type (e.g., vowels only) while keeping the 2k block pattern.
arrow_right_alt
Apply the same two-pointer invariant logic to an array of integers instead of a string.

help

常见问题

外企场景

继续练习

#556 下一个更大元素 III #557 反转字符串中的单词 III #524 通过删除字母匹配到字典里最长单词