[Swift]LeetCode204. 计数质数 | Count Primes

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Count the number of prime numbers less than a non-negative number, n.

Example:

Input: 10
Output: 4
Explanation: There are 4 prime numbers less than 10, they are 2, 3, 5, 7.

---

统计所有小于非负整数 n 的质数的数量。

示例:

输入: 10
输出: 4
解释: 小于 10 的质数一共有 4 个, 它们是 2, 3, 5, 7 。

---

88ms

class Solution {
    func countPrimes(_ n: Int) -> Int {
        guard n > 2 else { return 0 }
        var sieve = [Bool](repeating: true, count: n)
        var count = n / 2
        var i = 3
        
        while i * i < n {
            if sieve[i] {
                var j = i * i

                while j < n {
                    if sieve[j] {
                        count -= 1 
                        sieve[j] = false
                    }
                    j += 2 * i 
                }
            }
            i += 2
        }
        
        return count 
    }
}

---

356ms

class Solution{
    func countPrimes(_ n: Int) -> Int {
        if n < 3 { return 0 }
        var isPrime = Array.init(repeating: true, count: n)
        var i = 2;
        while i*i < n {
            if(!isPrime[i]) {
                i += 1;
                continue
            }
            var j = i*i
            while (j < n) {
                isPrime[j] = false
                j += i;
            }
            i += 1;
        }
        var counter = 0
        for k in 2..<n {
            if isPrime[k] {
                counter += 1
            }
        }
        return counter
    }
}

---

380ms

class Solution {
    func countPrimes(_ n: Int) -> Int {
        if n <= 0 {
            return 0
        }
        let newN = n - 1
        let sqrtN = Int(sqrt(Double(newN)))
        if n < 4 {
            return newN == 1 ? 0 : sqrtN
        }
        var nums = [Bool](repeating: true, count: n)

        var count = newN - 1
        for num in 0..<newN {
            nums[num] = true
        }
        nums[0] = false
        nums[1] = true
        
        for index in 2..<(sqrtN + 1){
            var j = index * index
            if nums[j - 1] {
                while j < n {
                    if nums[j - 1] {
                        count = count - 1
                        nums[j - 1] = false
                    }
                    j = j + index
                }
            }
        }
        
        return count
    }
}

---

408ms

class Solution {
    func countPrimes(_ n: Int) -> Int {        
        if n < 3 {
            return 0
        }
        if n == 3 {
            return 1
        }
        var array = [Bool](repeating: true, count: n)
        for i in 2...Int(sqrt(Double(n))) {
            if i * i >= n { break }
            if !array[i] { continue }
            var j = i * i
            while j < n {
                array[j] = false
                j = j + i
            }
        }
        var count = 0
        for i in 2..<n {
            if array[i] {
                count += 1
            }
        }
        return count
    }
}

---

416ms

class Solution 
{
    
    func countPrimes( _ n: Int ) -> Int 
    {
        var isPrime: [ Bool ] = Array< Bool >.init( repeating: true, count: max( 2, n+1 ) )
        
        isPrime[ 0 ] = false
        isPrime[ 1 ] = false
        var primeCount: Int = 0

        for primeVal in ( 0 ... n )
        {
            if isPrime[ primeVal ]
            {
                if primeVal < n
                { 
                    primeCount += 1
                }
                var factor: Int = 2 * primeVal
                while factor <= n
                {
                    isPrime[ factor ] = false
                    factor += primeVal
                }
            }
        }
        return primeCount
    }
}