[Swift]LeetCode970.强整数 | Powerful Integers

★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤微信公众号：山青咏芝（shanqingyongzhi）
➤博客园地址：山青咏芝（https://www.cnblogs.com/strengthen/）
➤GitHub地址：https://github.com/strengthen/LeetCode
➤原文地址：https://www.cnblogs.com/strengthen/p/10228288.html
➤如果链接不是山青咏芝的博客园地址，则可能是爬取作者的文章。
➤原文已修改更新！强烈建议点击原文地址阅读！支持作者！支持原创！
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★

热烈欢迎，请直接点击！！！

进入博主App Store主页，下载使用各个作品！！！

注：博主将坚持每月上线一个新app！！！

Given two non-negative integers x and y, an integer is powerful if it is equal to x^i + y^j for some integers i >= 0 and j >= 0.

Return a list of all powerful integers that have value less than or equal to bound.

You may return the answer in any order.  In your answer, each value should occur at most once.

Example 1:

Input: x = 2, y = 3, bound = 10
Output: [2,3,4,5,7,9,10]
Explanation: 
2 = 2^0 + 3^0
3 = 2^1 + 3^0
4 = 2^0 + 3^1
5 = 2^1 + 3^1
7 = 2^2 + 3^1
9 = 2^3 + 3^0
10 = 2^0 + 3^2

Example 2:

Input: x = 3, y = 5, bound = 15
Output: [2,4,6,8,10,14]

Note:

• 1 <= x <= 100
• 1 <= y <= 100
• 0 <= bound <= 10^6

---

给定两个非负整数 x 和 y，如果某一整数等于 x^i + y^j，其中整数 i >= 0 且 j >= 0，那么我们认为该整数是一个强整数。

返回值小于或等于 bound 的所有强整数组成的列表。

你可以按任何顺序返回答案。在你的回答中，每个值最多出现一次。

示例 1：

输入：x = 2, y = 3, bound = 10
输出：[2,3,4,5,7,9,10]
解释： 
2 = 2^0 + 3^0
3 = 2^1 + 3^0
4 = 2^0 + 3^1
5 = 2^1 + 3^1
7 = 2^2 + 3^1
9 = 2^3 + 3^0
10 = 2^0 + 3^2

示例 2：

输入：x = 3, y = 5, bound = 15
输出：[2,4,6,8,10,14]

提示：

• 1 <= x <= 100
• 1 <= y <= 100
• 0 <= bound <= 10^6

---

 8ms

class Solution {
    func powerfulIntegers(_ x: Int, _ y: Int, _ bound: Int) -> [Int] {
        var xs:[Int] = [1]
        var ys:[Int] = [1]
        
        if x > 1
        {
            var p:Int = x
            while(p <= bound)
            {
                xs.append(p)
                p *= x
            }
        }
        
        if y > 1
        {
            var p:Int = y
            while(p <= bound)
            {
                ys.append(p)
                p *= y
            }
        }
        
        var s:Set<Int> = Set<Int>()
        for xx in xs
        {
            for yy in ys
            {
                if xx + yy <= bound
                {
                    s.insert(xx + yy)
                }
            }
        }
        return Array(s)
    }
}

---

8ms

class Solution {
    func powerfulIntegers(_ x: Int, _ y: Int, _ bound: Int) -> [Int] {
        var res = [Int]()
        var xx = [Int]()
        var yy = [Int]()
        if x == 0 {
            xx = [0]
        } else if x == 1 {
            xx = [1]
        } else {
            xx.append(1)
            var tmp = x
            while tmp <= bound {
                xx.append(tmp)
                tmp *= x
            }
        }
        
        if y == 0 {
            yy = [0]
        } else if y == 1 {
            yy = [1]
        } else {
            yy.append(1)
            var tmp = y
            while tmp <= bound {
                yy.append(tmp)
                tmp *= y
            }
        }
        
        var tmp = 0
        
        for i in xx {
            for u in yy {
                tmp = i + u
                if !res.contains(tmp) && tmp <= bound {
                    res.append(tmp)
                }
                if tmp > bound {
                    break
                }
            }
        }
        
        return res
    }
}

---

16ms

class Solution {
    func powerfulIntegers(_ x: Int, _ y: Int, _ bound: Int) -> [Int] {
        var result = [Int]()
        helper(x, y, 0, 0, bound, &result)
        return result
    }
    
    func helper(_ x: Int, _ y: Int, _ powerX: Int, _ powerY: Int, _ bound: Int, _ result :inout [Int]) {
        let sum = Int(pow(Double(x), Double(powerX)) + pow(Double(y), Double(powerY)))
        if sum > bound {
            return
        }
        
        if !result.contains(sum) {
            result.append(sum)
        }
        
        if x > 1 {
            helper(x, y, powerX + 1, powerY, bound, &result)
        }
        
        if y > 1 {
            helper(x, y, powerX, powerY + 1, bound, &result)
        }
    }
}

---

20ms

class Solution {
    func powerfulIntegers(_ x: Int, _ y: Int, _ bound: Int) -> [Int] {
        if (x <= 0 && y <= 0) || (x == 1 && y == 0) || (x == 0 && y == 1) || (x == 1 && y == 1) {
            let a = Int(pow(Double(x), 0.0) + pow(Double(y), 0.0))
            let b = Int(pow(Double(x), 1.0) + pow(Double(y), 1.0))
            let c = Int(pow(Double(x), 1.0) + pow(Double(y), 0.0))
            let d = Int(pow(Double(x), 0.0) + pow(Double(y), 1.0))
            var s = Set<Int>()
            s.insert(a)
            s.insert(b)
            s.insert(c)
            s.insert(d)
            var arr = [Int]()
            for n in s {
                if n <= bound {
                    arr.append(n)
                }
            }
            return arr
        }
        var s = Set<Int>()
        var p1 = 0
        var sum = 0
        var count = 0
       out: while true {
            var p2 = 0
            while sum <= bound {
                sum = Int(pow(Double(x), Double(p1))) + Int(pow(Double(y), Double(p2)))
                p2 += 1
                if sum <= bound {
                    s.insert(sum)
                }
                count += 1
                if count > 3000 {
                    break out
                }
            }
            p1 += 1
            sum = Int(pow(Double(x), Double(p1))) + Int(pow(Double(y), Double(0)))
            if sum <= bound {
                
            } else {
                break
            }
        }
        return Array(s)
    }
}