[Swift]LeetCode692. 前K个高频单词 | Top K Frequent Words

时间 2019-11-11

标签 swift leetcode692 leetcode 高频单词 frequent words 栏目 Swift 繁體版

原文原文链接

★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤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： http://www.javashuo.com/article/p-wzrbftyx-me.html
➤若是连接不是山青咏芝的博客园地址，则多是爬取做者的文章。
➤原文已修改更新！强烈建议点击原文地址阅读！支持做者！支持原创！
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★html

Given a non-empty list of words, return the k most frequent elements.git

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.github

Example 1:微信

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:app

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:ide

You may assume k is always valid, 1 ≤ k ≤ number of unique elements.
Input words contain only lowercase letters.

Follow up:spa

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。code

返回的答案应该按单词出现频率由高到低排序。若是不一样的单词有相同出现频率，按字母顺序排序。htm

示例 1：blog

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" 以前。

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 }