README_EN.md

287. Find the Duplicate Number

中文文档

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.

 

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

 

Constraints:

• 2 <= n <= 3 * 104
• 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.

 

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)?

Solutions

Python3

class Solution:
    def findDuplicate(self, nums: List[int]) -> int:
        l, r = 0, len(nums) - 1
        while l < r:
            mid = (l + r) >> 1
            cnt = 0
            for e in nums:
                if e <= mid:
                    cnt += 1
            if cnt <= mid:
                l = mid + 1
            else:
                r = mid
        return l

Java

class Solution {
    public int findDuplicate(int[] nums) {
        int l = 1, r = nums.length - 1;
        while (l < r) {
            int mid = (l + r) >>> 1;
            int cnt = 0;
            for (int e : nums) {
                if (e <= mid) ++cnt;
            }
            if (cnt <= mid) l = mid + 1;
            else r = mid;
        }
        return l;
    }
}

C++

class Solution {
public:
    int findDuplicate(vector<int>& nums) {
        int l = 0, r = nums.size() - 1;
        while (l < r) {
            int mid = l + ((r - l) >> 1);
            int cnt = 0;
            for (auto e : nums) {
                if (e <= mid) ++cnt;
            }
            if (cnt <= mid) l = mid + 1;
            else r = mid;
        }
        return l;
    }
};

...