Given an integer array of size n, find all elements that appear more than ⌊ n/3 ⌋ times.
Note: The algorithm should run in linear time and in O(1) space.
Example 1:
Input: [3,2,3] Output: [3]
Example 2:
Input: [1,1,1,3,3,2,2,2] Output: [1,2]
给定一个大小为 n 的数组，找出其中所有出现超过 ⌊ n/3 ⌋ 次的元素。
说明: 要求算法的时间复杂度为 O(n)，空间复杂度为 O(1)。
示例 1:
输入: [3,2,3] 输出: [3]
示例 2:
输入: [1,1,1,3,3,2,2,2] 输出: [1,2]
96ms
1 class Solution { 2 func majorityElement(_ nums: [Int]) -> [Int] { 3 guard nums.count > 1 else { 4 return nums 5 } 6 7 var majorA = nums[0] 8 var countA = 0 9 10 var majorB = nums[0] 11 var countB = 0 12 13 for index in 0...(nums.count-1) { 14 if nums[index] == majorA { 15 countA += 1 16 continue 17 } 18 19 if nums[index] == majorB { 20 countB += 1 21 continue 22 } 23 24 if countA == 0 { 25 majorA = nums[index] 26 countA += 1 27 continue 28 } 29 30 if countB == 0 { 31 majorB = nums[index] 32 countB += 1 33 continue 34 } 35 36 countA -= 1 37 countB -= 1 38 } 39 40 countA = 0 41 countB = 0 42 for index in 0...(nums.count - 1) { 43 if nums[index] == majorA { 44 countA += 1 45 } else if nums[index] == majorB { 46 countB += 1 47 } 48 } 49 50 var result = [Int]() 51 if countA > nums.count/3 { 52 result.append(majorA) 53 } 54 if countB > nums.count/3 { 55 result.append(majorB) 56 } 57 58 return result 59 } 60 }
104ms
1 class Solution { 2 func majorityElement(_ nums: [Int]) -> [Int] { 3 4 var res = [Int]() 5 var cm = 0, cn = 0, m = 0, n = 0 6 for a in nums { 7 if a == m { cm += 1 } 8 else if a == n { cn += 1 } 9 else if cm == 0 { m = a; cm = 1} 10 else if cn == 0 { n = a; cn = 1 } 11 else { cm -= 1; cn -= 1} 12 } 13 cm = 0; cn = 0 14 for a in nums { 15 if a == m { cm += 1 } 16 else if a == n { cn += 1 } 17 } 18 if cm > nums.count / 3 {res.append(m)} 19 if cn > nums.count / 3 {res.append(n)} 20 return res 21 } 22 }
112ms
1 class Solution { 2 func majorityElement(_ nums: [Int]) -> [Int] { 3 var elems: [Int: Int] = [:] 4 for n in nums { 5 elems[n, default: 0] += 1 6 } 7 return elems.compactMap { key, value in 8 if value > nums.count / 3 { return key } 9 return nil 10 } 11 } 12 }
112ms
1 class Solution { 2 func majorityElement(_ nums: [Int]) -> [Int] { 3 var dict:[Int : Int] = [Int:Int]() 4 for num in nums { 5 if let n = dict[num] { 6 dict[num] = n + 1 7 } else { 8 dict[num] = 1 9 } 10 } 11 var r = [Int]() 12 let m = nums.count / 3 13 for (k,v) in dict { 14 if v > m { 15 r.append(k) 16 } 17 } 18 return r 19 } 20 }
116ms
1 class Solution { 2 func majorityElement(_ nums: [Int]) -> [Int] { 3 4 if nums.count == 0 { 5 return [] 6 } 7 8 var dict: [Int: Int] = [:] 9 var resultList: Set<Int> = [] 10 let limit = nums.count/3 11 12 for number in nums { 13 var result = 1 14 if let temp = dict[number] { 15 result = temp + 1 16 } 17 dict[number] = result 18 19 if result > limit { 20 resultList.insert(number) 21 } 22 } 23 24 return Array(resultList) 25 } 26 }
124ms
1 class Solution { 2 func majorityElement(_ nums: [Int]) -> [Int] { 3 4 if nums.count == 0 { 5 return [] 6 } 7 8 var dict: [Int: Int] = [:] 9 var resultList: [Int] = [] 10 let limit = nums.count/3 11 12 for number in nums { 13 var result = 1 14 if let temp = dict[number] { 15 result = temp + 1 16 } 17 dict[number] = result 18 19 if result > limit && !resultList.contains(number){ 20 resultList.append(number) 21 } 22 } 23 24 return resultList 25 } 26 }
140ms
1 class Solution { 2 func majorityElement(_ nums: [Int]) -> [Int] { 3 var dict = [Int: Int]() 4 var r = [Int]() 5 for n in nums { 6 dict[n, default: 0] += 1 7 if dict.keys.count == 3 { 8 for key in dict.keys { 9 dict[key, default: 0] -= 1 10 if dict[key, default: 0] == 0 { 11 dict[key] = nil 12 } 13 } 14 } 15 } 16 let c = nums.reduce(into: [Int: Int]()){ $0[$01, default: 0] += 1 } 17 return Array(dict.keys).filter { c[$0, default: 0] > nums.count / 3 } 18 } 19 }
156ms
1 class Solution { 2 func majorityElement(_ nums: [Int]) -> [Int] { 3 var res = [Int:Int].init() 4 5 for i in 0..<nums.count { 6 if res.keys.contains(nums[i]){ 7 let value = res[nums[i]]!+1 8 res[nums[i]] = value 9 }else{ 10 res[nums[i]] = 1 11 } 12 } 13 14 let dict = res.filter { (arg) -> Bool in 15 16 let (_, value) = arg 17 return value > nums.count/3 18 } 19 return Array(dict.keys) 20 } 21 }