What is the simplest approach to solve Shuffle String?

Use an auxiliary array of the same length as s and place each character at its target index from indices, then join into a string.

Can Shuffle String be solved in-place without extra space?

Yes, but you must track visited positions to avoid overwriting characters, which adds complexity.

What mistakes do people commonly make with indices mapping?

A frequent error is treating string positions and indices interchangeably, which can misplace characters.

How does GhostInterview improve solving array plus string problems?

It guides you to map elements correctly using indices and highlights potential overwrites or tracking errors for safe execution.

Are there variations of Shuffle String for interviews?

Yes, including cases with repeated characters, partial indices, or in-place space-optimized shuffling.

#1528

Easy

auto_awesome数组·string

LeetCode 题解工作台

重新排列字符串

给你一个字符串 s 和一个长度相同的整数数组 indices 。请你重新排列字符串 s ，其中第 i 个字符需要移动到 indices[i] 指示的位置。返回重新排列后的字符串。示例 1：输入： s = "codeleet", indices = [4,5,6,7,0,2,1,3] 输出…

数组字符串

题目描述

给你一个字符串 s 和一个 长度相同 的整数数组 indices 。

请你重新排列字符串 s ，其中第 i 个字符需要移动到 indices[i] 指示的位置。

返回重新排列后的字符串。

示例 1：

输入：s = "codeleet", indices = [4,5,6,7,0,2,1,3]
输出："leetcode"
解释：如图所示，"codeleet" 重新排列后变为 "leetcode" 。

示例 2：

输入：s = "abc", indices = [0,1,2]
输出："abc"
解释：重新排列后，每个字符都还留在原来的位置上。

提示：

s.length == indices.length == n
1 <= n <= 100
s 仅包含小写英文字母
0 <= indices[i] < n
indices 的所有的值都是唯一的

lightbulb

解题思路

方法一：模拟

我们创建一个与字符串长度相同的字符数组或字符串 $\textit{ans}$ ，然后遍历字符串 $\textit{s}$ ，将每个字符 $\textit{s}[i]$ 放置到 $\textit{ans}$ 的 $\textit{indices}[i]$ 位置上。最后将字符数组或字符串 $\textit{ans}$ 拼接成最终结果并返回。

时间复杂度 $O(n)$ ，空间复杂度 $O(n)$ ，其中 $n$ 是字符串的长度。

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class Solution:
    def restoreString(self, s: str, indices: List[int]) -> str:
        ans = [None] * len(s)
        for c, j in zip(s, indices):
            ans[j] = c
        return "".join(ans)

speed

复杂度分析

指标	值
时间	complexity is O(n) because each character is moved exactly once. Space complexity is O(n) if using an auxiliary array; in-place swaps can reduce space to O(1) but add bookkeeping overhead. The primary cost comes from handling the array mapping efficiently without collisions or overwrites.
空间	Depends on the final approach

psychology

面试官常问的追问

外企场景

question_mark
Do you recognize that direct mapping from indices is the simplest approach?
question_mark
Have you considered space-efficient in-place rearrangement versus auxiliary array?
question_mark
Can you handle edge cases like already sorted or reversed indices arrays?

warning

常见陷阱

外企场景

error
Overwriting characters without tracking positions leading to incorrect results.
error
Confusing indices positions with string positions during iteration.
error
Ignoring constraints on unique indices causing unexpected duplication or gaps.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Shuffle string with repeated characters and duplicate index handling.
arrow_right_alt
Shuffle string with partial indices and leaving some characters in place.
arrow_right_alt
Shuffle string in-place without using extra array to minimize space complexity.

help

常见问题

外企场景

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