287. Find the Duplicate Number

最新推荐文章于 2025-08-10 13:42:28 发布

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最新推荐文章于 2025-08-10 13:42:28 发布

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文章标签： runtime

原文链接：http://www.cnblogs.com/ordili/p/9991926.html

本文介绍了一种在未排序数组中查找重复数字的算法，该算法遵循LeetCode 287题的要求，不修改原始数组并使用常数额外空间。通过排序数组并遍历检查相邻元素来找出重复项，时间复杂度为O(Nlog(N))，空间复杂度为O(N)。

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1. Question:

287. Find the Duplicate Number

https://leetcode.com/problems/find-the-duplicate-number/

Given an array nums containing n + 1 integers where each integer is between 1 and n (inclusive), prove that at least one duplicate number must exist. Assume that there is only one duplicate number, find the duplicate one.

Example 1:

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

Example 2:

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

Note:

You must not modify the array (assume the array is read only).
You must use only constant, O(1) extra space.
Your runtime complexity should be less than O(n2).
There is only one duplicate number in the array, but it could be repeated more than once.

2. Solution:

class Solution:
    def findDuplicate(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        tmp = sorted(nums)
        bf = tmp[0]
        for each in tmp[1:]:
            if each == bf:
                return each
            bf = each

        return re