[Swift]LeetCode287. 寻找重复数 | Find the Duplicate Number

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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(n2).
4. There is only one duplicate number in the array, but it could be repeated more than once.

---

给定一个包含 n + 1 个整数的数组 nums，其数字都在 1 到 n 之间（包括 1 和 n），可知至少存在一个重复的整数。假设只有一个重复的整数，找出这个重复的数。

示例 1:

输入: [1,3,4,2,2]
输出: 2

示例 2:

输入: [3,1,3,4,2]
输出: 3

说明：

1. 不能更改原数组（假设数组是只读的）。
2. 只能使用额外的 O(1) 的空间。
3. 时间复杂度小于 O(n2) 。
4. 数组中只有一个重复的数字，但它可能不止重复出现一次。

---

32ms

class Solution {
    func findDuplicate(_ nums: [Int]) -> Int {
        var l = 1
        var r = nums.count - 1
        while l < r {
            let mid = (l + r) / 2
            var count = 0
            for i in 0 ..< nums.count {
                if nums[i] <= mid {
                    count += 1
                }
            }
            if count <= mid {
                l = mid + 1
            } else {
                r = mid
            }
        }
        return l
    }
}

---

52ms

class Solution {
    func findDuplicate(_ nums: [Int]) -> Int {
        var sorted = nums.sorted()
        var i = 1
        while i < sorted.count {
            if sorted[i] == sorted[i - 1] {
                return sorted[i]
            }
            i += 1
        }
        return -1
    }
}

---

80ms

class Solution {
    func findDuplicate(_ nums: [Int]) -> Int {
        var slow = 0, fast = 0, tmp = 0
        while true {
            slow = nums[slow]
            fast = nums[nums[fast]]
            if slow == fast {
                break
            }
        }
        while true {
            slow = nums[slow]
            tmp = nums[tmp]
            if slow == tmp {
                break
            }
        }
        
        return tmp
    }
}

---

92ms

class Solution {
    func findDuplicate(_ nums: [Int]) -> Int {
        var nums = nums
        
        for index in 0..<nums.count {
            
            while nums[index] != index {
                if nums[index] == nums[nums[index]] {
                    return nums[index]
                }
                nums.swapAt(index, nums[index])
            }
            
        }
        
        return 0
        
    }
}

---

96ms

class Solution {
    func findDuplicate(_ nums: [Int]) -> Int {
        var res = [Int:Int].init()
        
        for i in 0..<nums.count {
            if res.keys.contains(nums[i]) {
                return nums[i]
            }else{
                res[nums[i]] = i
            }
        }
        return 0
    }
}

---

120ms

class Solution {
    func findDuplicate(_ nums: [Int]) -> Int {
        
        var start = 0
        var end = nums.count
        
        while start < end {
            let mid = (start + end) / 2
            var count = 0
            
            nums.forEach{ if $0 <= mid { count += 1 } }
            
            if count > mid {
                end = mid
            }else{
                start = mid + 1
            }
        }

        return start
    }
}

---

124ms

class Solution {
    func findDuplicate(_ nums: [Int]) -> Int {
        var set = Set<Int>()
        
        for item in nums {
            if set.contains(item) {
                return item
            }else{
                set.update(with: item)
            }
        }
        
        return 0
    }
}

---

132ms

class Solution {
    func findDuplicate(_ nums: [Int]) -> Int {
        var set = Set<Int>()
        for (index, num) in nums.enumerated() {
            set.insert(num)
            if index == set.count {
                return num
            }
        }
        return 0
    }
}