[Swift]LeetCode884. 两句话中的不常见单词 | Uncommon Words from Two Sentences

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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:

1. 0 <= A.length <= 200
2. 0 <= B.length <= 200
3. A 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：

1. 0 <= A.length <= 200
2. 0 <= B.length <= 200
3. A 和 B 都只包含空格和小写字母。

---

8ms

class Solution {
    func uncommonFromSentences(_ A: String, _ B: String) -> [String] {
        var occurences = [String: Int]()
        let addSubstringToOccurences = { (substring: Substring) -> Void in
            let word = String(substring)
            occurences[word] = (occurences[word] ?? 0) + 1
        }

        A.split(separator: " ").forEach(addSubstringToOccurences)
        B.split(separator: " ").forEach(addSubstringToOccurences)

        return Array(occurences.filter { $1 == 1 }.keys)
    }
}

---

12ms

class Solution {
    func uncommonFromSentences(_ A: String, _ B: String) -> [String] {
        var result : [String] = []
        var DictA : [String:Int] = [:]
        var DictB : [String:Int] = [:]
        var wordsA = A.components(separatedBy: " ")
        var wordsB = B.components(separatedBy: " ")
        
        for word in wordsA {
            if var count = DictA[word] {
                DictA[word] = count + 1
            } else {
                DictA[word] = 1
            }
        }
        
        for word in wordsB {
            if var count = DictB[word] {
                DictB[word] = count + 1
            } else {
                DictB[word] = 1
            }
        }
        
        for key in DictA.keys {
            if DictA[key] == 1 && DictB[key] == nil {
                result.append(key)
            }
        }
        
        for key in DictB.keys {
            if DictB[key] == 1 && DictA[key] == nil {
                result.append(key)
            }
        }
        
        return result
    }
}

---

16ms

class Solution {
   func uncommonFromSentences(_ A: String, _ B: String) -> [String] {
        let str = A + " " + B
        var dic = Dictionary<String,Int>()
        var result = [String]()
        let subStr = str.split(separator: " ")
        let subs = subStr.map { String($0)}
        for sub in subs {
            if nil == dic[sub] {
                dic[sub] = 1
            } else {
                dic[sub] = dic[sub]! + 1
            }
        }
        dic.filter { (arg0) -> Bool in
            
            let (key, value) = arg0
            if value == 1 {
                result.append(key)
                return true
            } else {
                return false
            }
        }
        return result
    }
}

---

Runtime: 20 ms

Memory Usage: 20.3 MB

class Solution {
    func uncommonFromSentences(_ A: String, _ B: String) -> [String] {
        var count:[String:Int] = [String:Int]()
        var arr:[String] = (A + " " + B).components(separatedBy:" ")
        for w in arr
        {
            count[w,default:0] += 1
        }
        var res:[String] = [String]()
        for w in count.keys
        {
            if count[w,default:0] == 1
            {
                res.append(w)
            }
        }
        return res
    }
}

---

24ms

class Solution {
    func uncommonFromSentences(_ A: String, _ B: String) -> [String] {
        
        var wordCountDict: [Substring : Int] = [:]
        
        for word in A.split(separator: " ") {
             wordCountDict[word] = (wordCountDict[word] ?? 0) + 1
        }
        
        for word in B.split(separator: " ") {
            wordCountDict[word] = (wordCountDict[word] ?? 0) + 1
        }
        
        var answer: [String] = []
        
        for (word, count) in wordCountDict {
            if count == 1 {
                answer.append(String(word))
            }
        }
        
        return answer
    }
}

---

32ms

class Solution {
    func uncommonFromSentences(_ A: String, _ B: String) -> [String] {
        let words = A.split(separator: " ").map { String($0) } + B.split(separator: " ").map { String($0) }
        var countDict = [String: Int]()

        for word in words {
            countDict[word] = (countDict[word] ?? 0) + 1
        }

        return countDict.filter { $0.value == 1 }.map { $0.key }
    }
}

---

36ms

class Solution {
    func uncommonFromSentences(_ A: String, _ B: String) -> [String] {
        var result: [String] = []
        var count: [Substring: Int] = [:]
        let arrayA = A.split(separator: " ")
        let arrayB = B.split(separator: " ")
        for s in arrayA {
            if let value = count[s] {
                count[s] = value + 1
            } else {
                count[s] = 1
            }
        }
        for s in arrayB {
            if let value = count[s] {
                count[s] = value + 1
            } else {
                count[s] = 1
            }
        }
        for (key,value) in count {
            if value == 1 {
                result.append(String(key))
            }
        }
        return result
    }
}