287. Find the Duplicate Number

描述

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:

1. You must not modify the array (assume the array is read only).
2. You must use only constant, O(1) extra space.
3. Your runtime complexity should be less than O(n^2).
4. There is only one duplicate number in the array, but it could be repeated more than once.

解法

1)先排序，然后遍历一次，如果有重复的数字一定会相邻出现。

class Solution:
    def findDuplicate(self, nums: List[int]) -> int:
        nums.sort()
        for i in range(1, len(nums)):
            if nums[i] == nums[i - 1]:
                return nums[i]

排序的时间复杂度是 $O(NlogN)$。

2)使用集合记录重复元素。

class Solution:
    def findDuplicate(self, nums: List[int]) -> int:
        seen = {}
        for num in nums:
            if num in seen:
                return num
            seen.add(num)

时间复杂度 $O(n)$，空间复杂度 $O(n)$。

3)快慢指针，参考寻找环形链表的入口节点。

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

        ptr1 = nums[0]
        ptr2 = slow
        while ptr1 != ptr2:
            ptr1 = nums[ptr1]
            ptr2 = nums[ptr2]

        return ptr1

时间复杂度 $O(n)$，空间复杂度 $O(1)$。

引用