牛客网算法题：给定一个字符串，计算从做到右的字符出现的个数

题目：给定一个字符串，计算出从做到右的字符出现的个数

忽略字符计算后个数为1的数字

例如
原始输入字符串："​​​aabccccaaa​​​"
期望输出： “​​​a2bc4a3​​”

解释：
从左到右，

• a出现了2次，记为a2；
• b出现了1次，记为b；
• c出现了4次，记为c4；
• 再a出现了3次，记为a3。
• 结果就是：​​a2bc4a3​​

以Swift语言实现如下

class TestA: NSObject {

    private var lowerLetters: [String: Int] = ["a":0, "b":0, "c":0, "d":0, "e":0, "f":0, "g":0, "h":0, "i":0, "j":0, "k":0, "l":0, "m":0, "n":0, "o":0, "p":0, "q":0, "r":0, "s":0, "t":0, "u":0, "v":0, "w":0, "x":0, "y":0, "z":0]
    
    override init() {
        super.init()
        
        // 1.输入
        let s1 = "aabccccaaa"
        let s2 = "a2bc4a3"
        
        // 2.计算
        let result = self.calculate(with: s1)
        
        // 3.测试
        print("原始字符串      s1: \(s1)")
        print("原始期望结果     s2: \(s2)")
        print("计算后字符串 result: \(result)")
        if result == s2 {
            print("测试通过！")
        }
    }
    
    // 计算字符的个数
    private func calculate(with plainText: String) -> String {
        var finalStr: String = ""
        var lastCharacter: String = ""
        var currentIndex: Int = 0
        for c in plainText {
            if lastCharacter == String(c) {
                // 如果字符相等，继续加1
                let cCount: Int = lowerLetters[String(c)] ?? 0
                lowerLetters[String(c)] = cCount + 1
                
                if currentIndex == plainText.count - 1 {
                    // 如果是字符串的最后一个字符，则直接计算
                    finalStr += self.combineNumberOfOcurrencesCharacters(withLastCharacter: lastCharacter)
                }
            } else {
                // 如果不相等，就累计上一次字符的计算个数
                finalStr += self.combineNumberOfOcurrencesCharacters(withLastCharacter: lastCharacter)
                
                // 再把上一个字符置为0，否则后续出现的字符数量不对
                lowerLetters[lastCharacter] = 0
                
                // 计算当前字符的个数
                let cCount: Int = lowerLetters[String(c)] ?? 0
                lowerLetters[String(c)] = cCount + 1
            }
            currentIndex += 1
            lastCharacter = String(c)
        }
        return finalStr
    }
    
    // 组合字符出现的次数
    private func combineNumberOfOcurrencesCharacters(withLastCharacter lastCharacter: String) -> String {
        let lastLetterCount: Int = lowerLetters[lastCharacter] ?? 0
        if lastLetterCount > 1 {
            return "\(lastCharacter)\(lastLetterCount)"
        } else {
            return "\(lastCharacter)"
        }
    }   
    
}

输出测试结果如图

// 1.输入
let s1 = "aabccccaaa"
let s2 = "a2bc4a3"

// 2.计算
let result = self.calculate(with: s1)

// 3.测试
print("原始字符串      s1: \(s1)")
print("原始期望结果     s2: \(s2)")
print("计算后字符串 result: \(result)")
if result == s2 {
    print("测试通过！")
}