Leetcode** 287. Find the Duplicate Number

Description: 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.

Link: 287. Find the Duplicate Number

Examples:

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 = [1,1]
Output: 1

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

Follow up:

• How can we prove that at least one duplicate number must exist in nums?
• Can you solve the problem without modifying the array nums?
• Can you solve the problem using only constant, O(1) extra space?
• Can you solve the problem with runtime complexity less than O(n2)?

思路: 我们可以return collections.Counter(nums).most_common(1)[0][0]，或者类似的dict()计数，但是都不符合O(1)空间复杂度的要求，虽然时间是O(n). 不可以改变原nums.也就是不可以排序。所以按照题解的意思，所有的在nums中的数字都在区间[1,n]中，[1,n]共有n个unique numbers，如果一共有n+1个数，必然至少有一个数被重复了一次或者多次，题目中说，All the integers in nums appear only once except for precisely one integer which appears two or more times.只有一个数字出现两次或者多次，我们二分查找区间[1,n],而不是nums.具体地，l=1,r=n, 求mid，然后数一下nums中小于等于mid的个数，如果多余mid个，说明被重复的数在[1,mid]之间，反之，[mid+1,r]，注意结束条件是l < r，如果l<=r就会陷入死循环。时间复杂度是O(nlogn)

class Solution(object):
    def findDuplicate(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        l, r = 1, len(nums)-1
        while l < r:
            mid = int((l+r)/2)
            s = 0
            for i in nums:
                if i <= mid:
                    s += 1
            if s > mid:
                r = mid
            else:
                l = mid+1
        return l

看到还有O(n)时间复杂度的解法。

日期: 2021-04-14