[leetcode] 287. Find the Duplicate Number @ python

原创于 2019-02-22 07:13:24 发布 · 108 阅读

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本文探讨了在包含n+1个整数的数组中寻找重复数字的方法，每个整数介于1到n之间，确保只存在一个重复数字。文章提供了两种解决方案：使用集合和字典，详细解释了如何在不修改原始数组的情况下，利用O(1)额外空间找到重复数字，同时保持运行时间复杂度低于O(n^2)。

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原题

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.

解法1

集合

代码

class Solution:
    def findDuplicate(self, nums: 'List[int]') -> 'int':
        for n in set(nums):
            if nums.count(n) > 1:
                return n

解法2

字典

代码

class Solution:
    def findDuplicate(self, nums: 'List[int]') -> 'int':
        count = collections.Counter(nums)
        for n in count:
            if count[n] > 1:
                return n