[Swift]LeetCode771. 宝石与石头 | Jewels and Stones

★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤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： https://www.cnblogs.com/strengthen/p/10536345.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！
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★

You're given strings J representing the types of stones that are jewels, and S representing the stones you have.  Each character in S is a type of stone you have.  You want to know how many of the stones you have are also jewels.

The letters in J are guaranteed distinct, and all characters in J and S are letters. Letters are case sensitive, so "a" is considered a different type of stone from "A".

Example 1:

Input: J = "aA", S = "aAAbbbb"
Output: 3

Example 2:

Input: J = "z", S = "ZZ"
Output: 0

Note:

• S and J will consist of letters and have length at most 50.
• The characters in J are distinct.

---

 给定字符串J 代表石头中宝石的类型，和字符串 S代表你拥有的石头。 S 中每个字符代表了一种你拥有的石头的类型，你想知道你拥有的石头中有多少是宝石。

J 中的字母不重复，J 和 S中的所有字符都是字母。字母区分大小写，因此"a"和"A"是不同类型的石头。

示例 1:

输入: J = "aA", S = "aAAbbbb"
输出: 3

示例 2:

输入: J = "z", S = "ZZ"
输出: 0

注意:

• S 和 J 最多含有50个字母。
•  J 中的字符不重复。

---

Runtime: 8 ms

Memory Usage: 19.3 MB

class Solution {
    func numJewelsInStones(_ J: String, _ S: String) -> Int {
        var jewels = [Character : Int]()
        J.forEach { (c) in
            jewels[c] = 1
        }

        return S.reduce(0, { result, c in
            if let _ = jewels[c] {
                return result + 1
            }
            return result
        })
    }
}

---

8ms

class Solution {
    func numJewelsInStones(_ J: String, _ S: String) -> Int {
        var count: Int = 0
        for character in S {
            if J.contains(character) {
                count += 1
            }
        }
        return count
    }
}

---

12ms

class Solution {
    func numJewelsInStones(_ J: String, _ S: String) -> Int {
        return S.characters.filter{
            J.characters.contains($0)
        }.count
    }
}

---

16ms

class Solution {
    func numJewelsInStones(_ J: String, _ S: String) -> Int {
        var res = 0
        for x in J{
            for y in S{
                if(y==x){
                    res+=1
                }
            } 
        }  
        return res
    }
}

---

20ms

class Solution {
    func numJewelsInStones(_ J: String, _ S: String) -> Int {
        var matchArray: [Character] = []
        var stoneArray = Array(S)
        
        for jewel in J {
            if stoneArray.contains(jewel) {
                matchArray += stoneArray.filter { $0 == jewel }
            }
        }
        
        return matchArray.count
    }
}

---

19064 kb

class Solution {
    func numJewelsInStones(_ J: String, _ S: String) -> Int {
    var num = 0
    J.forEach { (Jcharacter) in
        S.forEach({ (Scharacter) in
            if Jcharacter == Scharacter {
                num += 1
            }
        })
    }
    
    return num
    }
}

---

48ms

class Solution {
    func numJewelsInStones(_ J: String, _ S: String) -> Int {
        var map: [Character:Int?] = [:]
        for str in J {
            map[str] = 0
        }
        for str in S {
            if let value = map[str] {
                map[str] = (value ?? 0) + 1
            }
        }
        return map.values.reduce(0) {$0 + ($1 ?? 0)}
    }
}