287. Find the Duplicate Number

环

Given an array of integers nums containing n + 1 integers where each integer is in the range [1, n] inclusive.

There is only one repeated number in nums, return this repeated number.

You must solve the problem without modifying the array nums and using only constant extra space.

Example 1:

Input: nums = [1,3,4,2,2]
Output: 2

Example 2:

Input: nums = [3,1,3,4,2]
Output: 3

Example 3:

Input: nums = [3,3,3,3,3]
Output: 3

Constraints:

• 1 <= n <= 105

• nums.length == n + 1

• 1 <= nums[i] <= n

• All the integers in nums appear only once except for precisely one integer which appears two or more times.

分析

1. 关键观察：

   • 将数组索引和值看作链表节点：

     • 索引 i 指向 nums[i]

     • 例如 nums = [1,3,4,2,2] 可转化为：

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       0 → 1 → 3 → 2 → 4 → 2 → 4 → ...

2. 重复数必然形成环：

   • 因为有重复数字，必然存在至少两个索引指向同一个值

   • 这会导致链表出现环（如示例中 2 → 4 → 2 的环）

class Solution:
    def findDuplicate(self, nums: List[int]) -> int:
        slow, fast = 0, 0
        while True:
            slow = nums[slow]
            fast = nums[nums[fast]]
            if slow == fast:
                break
        slow = 0
        while slow != fast:
            slow = nums[slow]
            fast = nums[fast]
        return slow