[Swift]LeetCode970.强整数|PowerfulIntegers

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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.

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 = 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 = 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 }