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Find Duplicate in Array ​

Problem Description ​

Given a read-only array of n + 1 integers between 1 and n, find one number that repeats in linear time using less than O(n) space and traversing the stream sequentially O(1) times.

If there are multiple possible answers ( like in the sample case ), output any one, if there is no duplicate, output -1

WARNING

no duplicate case, conflused? How can it happend

Problem Constraints

1 <= |A| <= 10^5
1 <= Ai <= |A|
1 <= |A| <= 10^5
1 <= Ai <= |A|

Input Format

The first argument is an integer array A.
The first argument is an integer array A.

Output Format

Return the integer that repeats in array A
Return the integer that repeats in array A

Example Input

Input 1:
A = [3, 4, 1, 4, 2]
Input 2:
A = [1, 2, 3]
Input 3:
A = [3, 4, 1, 4, 1]
Input 1:
A = [3, 4, 1, 4, 2]
Input 2:
A = [1, 2, 3]
Input 3:
A = [3, 4, 1, 4, 1]

Example Output

Output 1:
4
Output 2:
-1
Output 3:
1
Output 1:
4
Output 2:
-1
Output 3:
1

Example Explanation

Explanation 1:
4 repeats itself in the array [3, 4, 1, 4, 2]
Explanation 2:
No number repeats itself in the array [1, 2, 3]
Explanation 3:
1 and 4 repeats itself in the array [3, 4, 1, 4, 1], we can return 1 or 4
Explanation 1:
4 repeats itself in the array [3, 4, 1, 4, 2]
Explanation 2:
No number repeats itself in the array [1, 2, 3]
Explanation 3:
1 and 4 repeats itself in the array [3, 4, 1, 4, 1], we can return 1 or 4

Solution ​

swift
import Foundation

class Solution {
	func repeatedNumber(_ nums: [Int]) -> Int {
        var test = [Int](repeating:0,count:nums.count)
        for v in nums {
            if test[v] > 0 {
                return v
            }
            test[v] = 1
        }
        return -1
        
	}
}
import Foundation

class Solution {
	func repeatedNumber(_ nums: [Int]) -> Int {
        var test = [Int](repeating:0,count:nums.count)
        for v in nums {
            if test[v] > 0 {
                return v
            }
            test[v] = 1
        }
        return -1
        
	}
}

References ​