计算字符串中出现的字母

你的第二次尝试没有实现你想要实现的。 zip(inputString, 0...) Int 每个字符到字符本身的索引。

secondBaseCounts 将

打电话 Dictionary(secondBaseCounts, uniquingKeysWith: +) 对与重复键关联的每个值求和,这意味着 frequencies 字典将是所有索引的总和,其中某个字符出现在字典中 inputString 而不是那个角色出现的次数。

var inputString = "test this string"

extension String {
    
    func countCharacterOccurances() -> Dictionary<String, Any> {
        var occuranceDict : [String : Int] = [:]
       
        for i in self {
            var count = 1
            if occuranceDict[String(i)] == nil  {
                occuranceDict[String(i)] = count
            }
            else {
                count = occuranceDict[String(i)] ?? 0
                count += 1
                occuranceDict[String(i)] = count
            }
        }
        return occuranceDict as Dictionary<String, Any>
    }
}
var characterOccuranceDict = inputString.countCharacterOccurances()

输出:[“h”:1,“t”:4,“i”:2,“g”:1,“s”:3,“r”:1“,”:2,“n”:1,“e”:1]

let sortedDictionary = characterOccuranceDict.sorted { $0.key < $1.key }
sortedDictionary.forEach { (item) in 
    finalStr = finalStr + item.key + String(describing: item.value)
}
print(finalStr)

输出:A3B2C4D1

swift ios