Given a non-empty list of words, return the k most frequent elements.
Your answer should be sorted by frequency from highest to lowest. If two words have the same frequency, then the word with the lower alphabetical order comes first.
Example 1:
Input: ["i", "love", "leetcode", "i", "love", "coding"], k = 2
Output: ["i", "love"]
Explanation: "i" and "love" are the two most frequent words.
Note that "i" comes before "love" due to a lower alphabetical order.
Example 2:
Input: ["the", "day", "is", "sunny", "the", "the", "the", "sunny", "is", "is"], k = 4
Output: ["the", "is", "sunny", "day"]
Explanation: "the", "is", "sunny" and "day" are the four most frequent words,
with the number of occurrence being 4, 3, 2 and 1 respectively.
Note:
- You may assume k is always valid, 1 ≤ k ≤ number of unique elements.
- Input words contain only lowercase letters.
Follow up:
- Try to solve it in O(n log k) time and O(n) extra space.
给一非空的单词列表，返回前 k 个出现次数最多的单词。
返回的答案应该按单词出现频率由高到低排序。如果不同的单词有相同出现频率，按字母顺序排序。
示例 1：
输入: ["i", "love", "leetcode", "i", "love", "coding"], k = 2
输出: ["i", "love"]
解析: "i" 和 "love" 为出现次数最多的两个单词，均为2次。
注意，按字母顺序 "i" 在 "love" 之前。
示例 2：
输入: ["the", "day", "is", "sunny", "the", "the", "the", "sunny", "is", "is"], k = 4
输出: ["the", "is", "sunny", "day"]
解析: "the", "is", "sunny" 和 "day" 是出现次数最多的四个单词，
出现次数依次为 4, 3, 2 和 1 次。
注意：
- 假定 k 总为有效值， 1 ≤ k ≤ 集合元素数。
- 输入的单词均由小写字母组成。
扩展练习：
- 尝试以 O(n log k) 时间复杂度和 O(n) 空间复杂度解决。
76ms
1 class Solution { 2 func topKFrequent(_ words: [String], _ k: Int) -> [String] { 3 var frequencyList = [String:Int]() 4 for word in words { 5 if let count = frequencyList[word] { 6 frequencyList[word] = count + 1 7 } else { 8 frequencyList[word] = 1 9 } 10 } 11 let b = frequencyList.sorted { (dic1, dic2) -> Bool in 12 if dic1.value != dic2.value { 13 return dic1.value > dic2.value 14 } else { 15 return dic1.key < dic2.key 16 } 17 } 18 var result = [String]() 19 for item in b { 20 result.append(item.key) 21 if result.count == k { 22 return result 23 } 24 } 25 return result 26 } 27 }
Runtime: 88 ms
Memory Usage: 20.4 MB
1 class Solution { 2 func topKFrequent(_ words: [String], _ k: Int) -> [String] { 3 var k = k 4 var res:[String] = [String]() 5 var freq:[String:Int] = [String:Int]() 6 var v:[Set<String>] = [Set<String>](repeating:Set<String>(),count:words.count + 1) 7 for word in words 8 { 9 freq[word,default:0] += 1 10 } 11 for (key,val) in freq 12 { 13 v[val].insert(key) 14 } 15 for i in stride(from:v.count - 1,through:0,by:-1) 16 { 17 if k <= 0 {break} 18 var t:[String] = v[i].sorted() 19 if k >= t.count 20 { 21 res += t 22 } 23 else 24 { 25 26 res += t[0..<k] 27 } 28 k -= t.count 29 } 30 return res 31 } 32 }
88ms
1 class Solution { 2 func topKFrequent(_ words: [String], _ k: Int) -> [String] { 3 var countMap = [String: Int]() 4 var firstAppearance = [String: Int]() 5 for (i, word) in words.enumerated() { 6 countMap[word, default: 0] += 1 7 if firstAppearance[word] == nil { 8 firstAppearance[word] = i 9 } 10 } 11 var sortedStringsBasedOnCount = countMap.sorted { 12 return $0.value > $1.value || 13 ($0.value == $1.value && $0.key < $1.key) 14 } 15 return sortedStringsBasedOnCount[0..<k].map { $0.0 } 16 } 17 }
92ms
1 class Solution { 2 func topKFrequent(_ words: [String], _ k: Int) -> [String] { 3 var countMap = [String: Int]() 4 words.forEach { countMap[$0, default: 0] += 1 } 5 var sortedStringsBasedOnCount = countMap.sorted { 6 return $0.value > $1.value || 7 ($0.value == $1.value && $0.key < $1.key) 8 } 9 return sortedStringsBasedOnCount[0..<k].map { $0.0 } 10 } 11 }
96ms
1 class Solution { 2 func topKFrequent(_ words: [String], _ k: Int) -> [String] { 3 var wordDict = [String : Int]() 4 for word in words { 5 wordDict[word] = 1 + (wordDict[word] ?? 0) 6 } 7 return wordDict.sorted { 8 $0.value > $1.value || $0.value == $1.value && $0.key < $1.key 9 }[0..<k].map{$0.key} 10 } 11 }
100ms
1 class Solution { 2 func topKFrequent(_ words: [String], _ k: Int) -> [String] { 3 var frequencyList = [String:Int]() 4 for word in words { 5 if let count = frequencyList[word] { 6 frequencyList[word] = count + 1 7 } else { 8 frequencyList[word] = 1 9 } 10 } 11 var frequencyList2 = [[String]](repeating:[String](), count:words.count) 12 for (key, value) in frequencyList { 13 frequencyList2[value].append(key) 14 frequencyList2[value] = frequencyList2[value].sorted() 15 } 16 frequencyList2 = frequencyList2.filter{$0 != [String]()} 17 var result = [String]() 18 var index = frequencyList2.count - 1 19 while index >= 0 { 20 for item in frequencyList2[index] { 21 result.append(item) 22 if result.count == k { 23 return result 24 } 25 } 26 index -= 1 27 } 28 return result 29 } 30 }
128ms
1 class Solution { 2 func topKFrequent(_ words: [String], _ k: Int) -> [String] { 3 guard words.count > 0 else { 4 return [String]() 5 } 6 7 var memo = [String:Int]() 8 9 for w in words { 10 memo[w] = memo[w, default:0] + 1 11 } 12 13 var helpArray = [[String]]() 14 15 for key in memo.keys { 16 if let count = memo[key] { 17 helpArray.append([String(count), key]) 18 } 19 } 20 21 helpArray.sort(by:{ 22 if Int($0[0])! > Int($1[0])! { 23 return true 24 } else if Int($0[0])! < Int($1[0])! { 25 return false 26 } else { 27 return $0[1] < $1[1] 28 } 29 }) 30 var sortedArray = helpArray.map{ $0[1] } 31 32 return Array(sortedArray[0...k - 1]) 33 } 34 }