网站维护中一般要多长时间,wordpress 分享到微信二维码,一站式网站建设费用,学服装设计的就业方向★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★➤微信公众号#xff1a;山青咏芝#xff08;shanqingyongzhi#xff09;➤博客园地址#xff1a;山青咏芝#xff08;https://www.cnblogs.com/strengthen/#xff09;➤GitHub地址山青咏芝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 https://www.cnblogs.com/strengthen/p/10603493.html ➤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★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★ We are given two sentences A and B. (A sentence is a string of space separated words. Each word consists only of lowercase letters.) A word is uncommon if it appears exactly once in one of the sentences, and does not appear in the other sentence. Return a list of all uncommon words. You may return the list in any order. Example 1: Input: A this apple is sweet, B this apple is sour
Output: [sweet,sour]Example 2: Input: A apple apple, B banana
Output: [banana] Note: 0 A.length 2000 B.length 200A and B both contain only spaces and lowercase letters.给定两个句子 A 和 B 。 句子是一串由空格分隔的单词。每个单词仅由小写字母组成。 如果一个单词在其中一个句子中只出现一次在另一个句子中却没有出现那么这个单词就是不常见的。 返回所有不常用单词的列表。 您可以按任何顺序返回列表。 示例 1 输入A this apple is sweet, B this apple is sour
输出[sweet,sour]示例 2 输入A apple apple, B banana
输出[banana] 提示 0 A.length 2000 B.length 200A 和 B 都只包含空格和小写字母。8ms 1 class Solution {2 func uncommonFromSentences(_ A: String, _ B: String) - [String] {3 var occurences [String: Int]()4 let addSubstringToOccurences { (substring: Substring) - Void in5 let word String(substring)6 occurences[word] (occurences[word] ?? 0) 17 }8 9 A.split(separator: ).forEach(addSubstringToOccurences)
10 B.split(separator: ).forEach(addSubstringToOccurences)
11
12 return Array(occurences.filter { $1 1 }.keys)
13 }
14 } 12ms 1 class Solution {2 func uncommonFromSentences(_ A: String, _ B: String) - [String] {3 var result : [String] []4 var DictA : [String:Int] [:]5 var DictB : [String:Int] [:]6 var wordsA A.components(separatedBy: )7 var wordsB B.components(separatedBy: )8 9 for word in wordsA {
10 if var count DictA[word] {
11 DictA[word] count 1
12 } else {
13 DictA[word] 1
14 }
15 }
16
17 for word in wordsB {
18 if var count DictB[word] {
19 DictB[word] count 1
20 } else {
21 DictB[word] 1
22 }
23 }
24
25 for key in DictA.keys {
26 if DictA[key] 1 DictB[key] nil {
27 result.append(key)
28 }
29 }
30
31 for key in DictB.keys {
32 if DictB[key] 1 DictA[key] nil {
33 result.append(key)
34 }
35 }
36
37 return result
38 }
39 } 16ms 1 class Solution {2 func uncommonFromSentences(_ A: String, _ B: String) - [String] {3 let str A B4 var dic DictionaryString,Int()5 var result [String]()6 let subStr str.split(separator: )7 let subs subStr.map { String($0)}8 for sub in subs {9 if nil dic[sub] {
10 dic[sub] 1
11 } else {
12 dic[sub] dic[sub]! 1
13 }
14 }
15 dic.filter { (arg0) - Bool in
16
17 let (key, value) arg0
18 if value 1 {
19 result.append(key)
20 return true
21 } else {
22 return false
23 }
24 }
25 return result
26 }
27 } Runtime: 20 ms Memory Usage: 20.3 MB 1 class Solution {2 func uncommonFromSentences(_ A: String, _ B: String) - [String] {3 var count:[String:Int] [String:Int]()4 var arr:[String] (A B).components(separatedBy: )5 for w in arr6 {7 count[w,default:0] 18 }9 var res:[String] [String]()
10 for w in count.keys
11 {
12 if count[w,default:0] 1
13 {
14 res.append(w)
15 }
16 }
17 return res
18 }
19 } 24ms 1 class Solution {2 func uncommonFromSentences(_ A: String, _ B: String) - [String] {3 4 var wordCountDict: [Substring : Int] [:]5 6 for word in A.split(separator: ) {7 wordCountDict[word] (wordCountDict[word] ?? 0) 18 }9
10 for word in B.split(separator: ) {
11 wordCountDict[word] (wordCountDict[word] ?? 0) 1
12 }
13
14 var answer: [String] []
15
16 for (word, count) in wordCountDict {
17 if count 1 {
18 answer.append(String(word))
19 }
20 }
21
22 return answer
23 }
24 } 32ms 1 class Solution {2 func uncommonFromSentences(_ A: String, _ B: String) - [String] {3 let words A.split(separator: ).map { String($0) } B.split(separator: ).map { String($0) }4 var countDict [String: Int]()5 6 for word in words {7 countDict[word] (countDict[word] ?? 0) 18 }9
10 return countDict.filter { $0.value 1 }.map { $0.key }
11 }
12 } 36ms 1 class Solution {2 func uncommonFromSentences(_ A: String, _ B: String) - [String] {3 var result: [String] []4 var count: [Substring: Int] [:]5 let arrayA A.split(separator: )6 let arrayB B.split(separator: )7 for s in arrayA {8 if let value count[s] {9 count[s] value 1
10 } else {
11 count[s] 1
12 }
13 }
14 for s in arrayB {
15 if let value count[s] {
16 count[s] value 1
17 } else {
18 count[s] 1
19 }
20 }
21 for (key,value) in count {
22 if value 1 {
23 result.append(String(key))
24 }
25 }
26 return result
27 }
28 } 转载于:https://www.cnblogs.com/strengthen/p/10603493.html
