zoukankan      html  css  js  c++  java
  • [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!
    ★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★

    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

     1 class Solution {
     2     func powerfulIntegers(_ x: Int, _ y: Int, _ bound: Int) -> [Int] {
     3         var xs:[Int] = [1]
     4         var ys:[Int] = [1]
     5         
     6         if x > 1
     7         {
     8             var p:Int = x
     9             while(p <= bound)
    10             {
    11                 xs.append(p)
    12                 p *= x
    13             }
    14         }
    15         
    16         if y > 1
    17         {
    18             var p:Int = y
    19             while(p <= bound)
    20             {
    21                 ys.append(p)
    22                 p *= y
    23             }
    24         }
    25         
    26         var s:Set<Int> = Set<Int>()
    27         for xx in xs
    28         {
    29             for yy in ys
    30             {
    31                 if xx + yy <= bound
    32                 {
    33                     s.insert(xx + yy)
    34                 }
    35             }
    36         }
    37         return Array(s)
    38     }
    39 }

    8ms

     1 class Solution {
     2     func powerfulIntegers(_ x: Int, _ y: Int, _ bound: Int) -> [Int] {
     3         var res = [Int]()
     4         var xx = [Int]()
     5         var yy = [Int]()
     6         if x == 0 {
     7             xx = [0]
     8         } else if x == 1 {
     9             xx = [1]
    10         } else {
    11             xx.append(1)
    12             var tmp = x
    13             while tmp <= bound {
    14                 xx.append(tmp)
    15                 tmp *= x
    16             }
    17         }
    18         
    19         if y == 0 {
    20             yy = [0]
    21         } else if y == 1 {
    22             yy = [1]
    23         } else {
    24             yy.append(1)
    25             var tmp = y
    26             while tmp <= bound {
    27                 yy.append(tmp)
    28                 tmp *= y
    29             }
    30         }
    31         
    32         var tmp = 0
    33         
    34         for i in xx {
    35             for u in yy {
    36                 tmp = i + u
    37                 if !res.contains(tmp) && tmp <= bound {
    38                     res.append(tmp)
    39                 }
    40                 if tmp > bound {
    41                     break
    42                 }
    43             }
    44         }
    45         
    46         return res
    47     }
    48 }

    16ms

     1 class Solution {
     2     func powerfulIntegers(_ x: Int, _ y: Int, _ bound: Int) -> [Int] {
     3         var result = [Int]()
     4         helper(x, y, 0, 0, bound, &result)
     5         return result
     6     }
     7     
     8     func helper(_ x: Int, _ y: Int, _ powerX: Int, _ powerY: Int, _ bound: Int, _ result :inout [Int]) {
     9         let sum = Int(pow(Double(x), Double(powerX)) + pow(Double(y), Double(powerY)))
    10         if sum > bound {
    11             return
    12         }
    13         
    14         if !result.contains(sum) {
    15             result.append(sum)
    16         }
    17         
    18         if x > 1 {
    19             helper(x, y, powerX + 1, powerY, bound, &result)
    20         }
    21         
    22         if y > 1 {
    23             helper(x, y, powerX, powerY + 1, bound, &result)
    24         }
    25     }
    26 }

    20ms

     1 class Solution {
     2     func powerfulIntegers(_ x: Int, _ y: Int, _ bound: Int) -> [Int] {
     3         if (x <= 0 && y <= 0) || (x == 1 && y == 0) || (x == 0 && y == 1) || (x == 1 && y == 1) {
     4             let a = Int(pow(Double(x), 0.0) + pow(Double(y), 0.0))
     5             let b = Int(pow(Double(x), 1.0) + pow(Double(y), 1.0))
     6             let c = Int(pow(Double(x), 1.0) + pow(Double(y), 0.0))
     7             let d = Int(pow(Double(x), 0.0) + pow(Double(y), 1.0))
     8             var s = Set<Int>()
     9             s.insert(a)
    10             s.insert(b)
    11             s.insert(c)
    12             s.insert(d)
    13             var arr = [Int]()
    14             for n in s {
    15                 if n <= bound {
    16                     arr.append(n)
    17                 }
    18             }
    19             return arr
    20         }
    21         var s = Set<Int>()
    22         var p1 = 0
    23         var sum = 0
    24         var count = 0
    25        out: while true {
    26             var p2 = 0
    27             while sum <= bound {
    28                 sum = Int(pow(Double(x), Double(p1))) + Int(pow(Double(y), Double(p2)))
    29                 p2 += 1
    30                 if sum <= bound {
    31                     s.insert(sum)
    32                 }
    33                 count += 1
    34                 if count > 3000 {
    35                     break out
    36                 }
    37             }
    38             p1 += 1
    39             sum = Int(pow(Double(x), Double(p1))) + Int(pow(Double(y), Double(0)))
    40             if sum <= bound {
    41                 
    42             } else {
    43                 break
    44             }
    45         }
    46         return Array(s)
    47     }
    48 }
  • 相关阅读:
    javamail发送邮件
    java复制文件夹中的所有文件和文件夹到另一个文件夹中
    jsp中使用out和response.getOutputStream的方法
    PHP数组用法
    PHP中的session
    java中list按照某个属性排序方法
    java读取xml文件内容
    C#实现验证码
    java实现验证码功能
    Java IO(四)------字节输入输出流
  • 原文地址:https://www.cnblogs.com/strengthen/p/10228288.html
Copyright © 2011-2022 走看看