[Swift]LeetCode264.丑数 II | Ugly Number II

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Write a program to find the n-th ugly number.

Ugly numbers are positive numbers whose prime factors only include 2, 3, 5. 

Example:

Input: n = 10
Output: 12
Explanation: 1, 2, 3, 4, 5, 6, 8, 9, 10, 12 is the sequence of the first 10 ugly numbers.

Note:  

1. 1 is typically treated as an ugly number.
2. n does not exceed 1690.

---

编写一个程序，找出第 n 个丑数。

丑数就是只包含质因数 2, 3, 5 的正整数。

示例:

输入: n = 10
输出: 12
解释: 1, 2, 3, 4, 5, 6, 8, 9, 10, 12 是前 10 个丑数。

说明:  

1. 1 是丑数。
2. n 不超过1690。

---

16ms

class Solution {
    func nthUglyNumber(_ n: Int) -> Int {
        if n == 1 { return 1}
        var val1 = 2
        var val2 = 3
        var val3 = 5
        var i1 = 0
        var i2 = 0
        var i3 = 0
        var result = [1]
        for i in 1 ..< n {
            let current = min(min(val1, val2), val3)
            result.append(current)
            if current == val1 {
                i1 += 1
                val1 = result[i1] * 2
            }
            if current == val2 {
                i2 += 1
                val2 = result[i2] * 3
            }
            if current == val3 {
                i3 += 1
                val3 = result[i3] * 5
            }
        }
        return result[n - 1]
    }
}

---

20ms

class Solution {
    func nthUglyNumber(_ n: Int) -> Int{

        var res = [1]
        var i2 = 0, i3 = 0, i5 = 0
        while res.count < n {
            let m2 = res[i2]*2, m3 = res[i3]*3, m5 = res[i5]*5
            let mn = min(min(m2, m3), m5)
            if m2 == mn { i2 += 1 }
            if m3 == mn { i3 += 1 }
            if m5 == mn { i5 += 1 }
            res.append(mn)
        }
        return res.last!
    }
}

---

36ms

class Solution {
    func nthUglyNumber(_ n: Int) -> Int {
        var res = [1]
        var i = 0
        var j = 0
        var k = 0
        
        while res.count < n {
            
            res.append(res[i] * 2)
            i += 1

            while res[j] * 3 < res[i] * 2 || res[k] * 5 < res[i] * 2 {
                if res[j] * 3 < res[i] * 2 && res[j] * 3 < res[k] * 5 {
                    if res[j] % 2 != 0 {
                        res.append(res[j] * 3)
                    }
                    j += 1
                }
                else {
                    if res[k] % 2 != 0 && res[k] % 3 != 0 {
                        res.append(res[k] * 5)
                    }
                    k += 1
                }
            }
            
        }
        
        return res[n-1]
    }
}

---

44ms

class Solution {
    func nthUglyNumber(_ n: Int) -> Int {
        var res = [1]
        var c2 = 0, c3 = 0, c5 = 0
        var t2, t3, t5: Int
        var minp: Int
        
        for _ in 1 ..< n {
            t2 = res[c2] * 2
            t3 = res[c3] * 3
            t5 = res[c5] * 5
            minp = min(t2, t3, t5)
            if t2 == minp {
                c2 += 1
            }
            if t3 == minp {
                c3 += 1
            }
            if t5 == minp {
                c5 += 1
            }
            res.append(minp)
        }

        return res.last!
    }
}