[Swift]LeetCode137.只出现一次的数字II|SingleNumberII

OStack程序员社区-中国程序员成长平台 › 门户 › 编程› 移动开发›Swift 教程

原作者: [db:作者] 来自: [db:来源] 收藏邀请

★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤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：
➤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 a non-empty array of integers, every element appears three times except for one, which appears exactly once. Find that single one.

Note:

Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?

Example 1:

Input: [2,2,3,2]
Output: 3

Example 2:

Input: [0,1,0,1,0,1,99]
Output: 99

给定一个非空整数数组，除了某个元素只出现一次以外，其余每个元素均出现了三次。找出那个只出现了一次的元素。

说明：

你的算法应该具有线性时间复杂度。你可以不使用额外空间来实现吗？

示例 1:

输入: [2,2,3,2]
输出: 3

示例 2:

输入: [0,1,0,1,0,1,99]
输出: 99

12ms

 1 class Solution {
 2     func singleNumber(_ nums: [Int]) -> Int {
 3         let count = nums.count
 4         guard count > 0 else { return 0 }
 5         let range = 1..<count
 6         var oneTime = nums[0]
 7         var twoTimes = 0
 8         var notThreeTimes = 0
 9         for i in range {
10             let next = nums[i]
11             twoTimes |= oneTime & next
12             oneTime ^= next
13             notThreeTimes = ~(oneTime & twoTimes)
14             oneTime &= notThreeTimes
15             twoTimes &= notThreeTimes
16         }
17         return oneTime
18     }
19 }

16ms

 1 class Solution {
 2     func singleNumber(_ nums: [Int]) -> Int {
 3         var a = 0, b = 0;
 4         for num in nums {
 5             b = (b ^ num) & ~a;
 6             a = (a ^ num) & ~b;
 7         }
 8         return b;
 9     }
10 }

24ms

 1 class Solution {
 2     func singleNumber(_ nums: [Int]) -> Int {
 3         guard nums.count > 0 else {
 4             return 0
 5         }
 6         
 7         var temp = [Int: Int]()
 8         
 9         for num in nums {
10             temp[num] = temp[num, default: 0] + 1
11         }
12         
13         for key in temp.keys {
14             if temp[key]! == 1 {
15                 return key
16             } 
17         }
18         
19         return 0
20     }
21 }

28ms

 1 class Solution {
 2     func singleNumber(_ nums: [Int]) -> Int {
 3         var resSet = Set<Int>(nums)
 4         var showed = Set<Int>()
 5         for num in nums {
 6             if showed.contains(num) {
 7                 resSet.remove(num)
 8             }
 9             showed.insert(num)
10         }
11         return resSet.first!
12     }
13 }

76ms

 1 class Solution {
 2     func singleNumber(_ nums: [Int]) -> Int {
 3         var numDict = [Int: Int]()
 4         for num in nums {
 5             if let value = numDict[num] {
 6                 numDict[num] = value + 1
 7             } else {
 8                 numDict[num] = 1
 9             }
10         }
11             
12         let singleVal = numDict.filter { $0.value == 1 }.first!
13         return singleVal.key
14     }
15 }

80ms

1 class Solution {
2     func singleNumber(_ nums: [Int]) -> Int {
3         return ((Set(nums).reduce(0,+) * 3) - nums.reduce(0,+))/2
4     }
5 }

80ms

 1 class Solution {
 2     func singleNumber(_ nums: [Int]) -> Int {
 3         guard nums.count > 0 else {
 4             return 0
 5         }
 6         var result = 0
 7         for i in 0..<MemoryLayout<Int>.size * 8 {
 8             var count = 0
 9             for j in 0..<nums.count {
10                 count += (nums[j] >> i) & 1
11             }
12             
13             result += (count % 3) << i
14         }
15         
16         return result
17     }
18 }