[Swift]LeetCode970.强整数 | Powerful Integers

时间 2019-11-11

标签 swift leetcode970 leetcode 整数 powerful integers 栏目 Swift 繁體版

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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.git

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

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

Example 1:app

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:less

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

Note:spa

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

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

返回值小于或等于 bound 的全部强整数组成的列表。htm

你能够按任何顺序返回答案。在你的回答中，每一个值最多出现一次。blog

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 }