如何快速查找字符串数组中出现的字母答案

【问题标题】：How to find occurences of a lettter in the string array in swift如何快速查找字符串数组中出现的字母
【发布时间】：2021-11-05 04:33:31
【问题描述】：

我是尝试实现此逻辑的初学者，任何人都可以建议该逻辑。

func findLetterOccurence(Letter: String){
    let array = ["Data", "program", "questions", "Helpful"]

    ///Logic to find the given letter occurences in string array

    print("\(Letter) occured in \(count) times")
}

预期输出：a 出现了 3 次

我试过如下：

var count = 0
for i in array {
   var newArray.append(i)
   count = components(separatedBy: newArray).count - 1
}

但我不明白 components(separatedBy:) 内部的逻辑到底是什么？我的意思是如果没有更高的功能，我们怎么能在这里实现逻辑。

【问题讨论】：

你尝试了什么？
我已经更新了先生的问题，请看一下
对数组中的每个元素使用components(separatedBy:)。

标签： ios arrays swift objective-c xcode

【解决方案1】：

试试这样的：

 func findLetterOccurence(letter: String) {
    var count = 0
    for word in array { count += word.filter{ String($0) == letter}.count }
    print("--> \(letter) occured in \(count) times")
}

如果要不区分大小写，则必须进行调整，如下所示：

func findLetterOccurence(letter: String) {
    var count = 0
    for word in array { count += word.filter{ String($0).lowercased() == letter.lowercased()}.count }
    print("--> \(letter) occured in \(count) times")
}

【讨论】：

我也喜欢使用过滤器的想法

【解决方案2】：

添加此扩展以查找字符串中后者的出现

extension String {
    func numberOfOccurrencesOf(string: String) -> Int {
        return self.components(separatedBy:string).count - 1
    }
}

将它用于数组

        func findLetterOccurence(Letter: String){
        let array = ["Data", "program", "questions", "Helpful"]
        var number = 0
        for str in array{
            var l = Letter
//            uncomment it to use code for upper case and lower case both
//            l = str.lowercased()
            
            number = number + str.numberOfOccurrencesOf(string: l)
        }
        print("\(Letter) occured in \(number) times")
    }

【讨论】：

【解决方案3】：

几种方法。

    @discardableResult func findLetterOccurence(letter: String) -> Int {
    
    let array = ["Data", "program", "questions", "Helpful"]
    var count = 0
    // here we join the array into a single string. Then for each character we check if the lowercased version matches the string lowercased value. 
    array.joined().forEach({ if $0.lowercased() == letter.lowercased() { count += 1} } )
    
    print("\(letter) occured in \(count) times")
    
    return count
}

你也可以做一个敏感的比较，不用管外壳通过说

array.joined().forEach({ if String($0) == letter { count += 1} } )

另一种方式是这样

//here our argument is a character because maybe we just want to search for a single letter.
    @discardableResult func findLetterOccurence2(character: Character) -> Int {
    
        let array = ["Data", "program", "questions", "Helpful"]
   
    //again join the array into a single string. and reduce takes the `into` parameter and passes it into the closure as $0 in this case, and each element of the string gets passed in the second argument of the closure.
        let count = array.joined().reduce(into: 0) {
           $0 += $1.lowercased() == letter.lowercased() ? 1 : 0
        }
        print("\(letter) occured in \(count) times")
    
        return count
    }

【讨论】：