[Swift]LeetCode880. 索引处的解码字符串 | Decoded String at Index

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An encoded string S is given.  To find and write the decodedstring to a tape, the encoded string is read one character at a time and the following steps are taken:

• If the character read is a letter, that letter is written onto the tape.
• If the character read is a digit (say d), the entire current tape is repeatedly written d-1 more times in total.

Now for some encoded string S, and an index K, find and return the K-th letter (1 indexed) in the decoded string.

Example 1:

Input: S = "leet2code3", K = 10
Output: "o"
Explanation: 
The decoded string is "leetleetcodeleetleetcodeleetleetcode".
The 10th letter in the string is "o".

Example 2:

Input: S = "ha22", K = 5
Output: "h"
Explanation: 
The decoded string is "hahahaha".  The 5th letter is "h".

Example 3:

Input: S = "a2345678999999999999999", K = 1
Output: "a"
Explanation: 
The decoded string is "a" repeated 8301530446056247680 times.  The 1st letter is "a".

Note:

1. 2 <= S.length <= 100
2. S will only contain lowercase letters and digits 2through 9.
3. S starts with a letter.
4. 1 <= K <= 10^9
5. The decoded string is guaranteed to have less than 2^63 letters.

---

给定一个编码字符串 S。为了找出解码字符串并将其写入磁带，从编码字符串中每次读取一个字符，并采取以下步骤：

• 如果所读的字符是字母，则将该字母写在磁带上。
• 如果所读的字符是数字（例如 d），则整个当前磁带总共会被重复写 d-1 次。

现在，对于给定的编码字符串 S 和索引 K，查找并返回解码字符串中的第 K 个字母。 

示例 1：

输入：S = "leet2code3", K = 10
输出："o"
解释：
解码后的字符串为 "leetleetcodeleetleetcodeleetleetcode"。
字符串中的第 10 个字母是 "o"。

示例 2：

输入：S = "ha22", K = 5
输出："h"
解释：
解码后的字符串为 "hahahaha"。第 5 个字母是 "h"。

示例 3：

输入：S = "a2345678999999999999999", K = 1
输出："a"
解释：
解码后的字符串为 "a" 重复 8301530446056247680 次。第 1 个字母是 "a"。 

提示：

1. 2 <= S.length <= 100
2. S 只包含小写字母与数字 2 到 9 。
3. S 以字母开头。
4. 1 <= K <= 10^9
5. 解码后的字符串保证少于 2^63 个字母。

---

Runtime: 4 ms

Memory Usage: 19.6 MB

class Solution {
    func decodeAtIndex(_ S: String, _ K: Int) -> String {
        var K = K
        var N:Int = 0
        var arrS:[Character] = Array(S)
        var countS:Int = S.count < K ? S.count : K        
        for i in 0..<countS
        {
            let num:Int = arrS[i].ascii
            N = num >= 48 && num <= 57 ? N * (num - 48) : N + 1
        }
        var i = countS - 1
        while (true)
        {
            let num:Int = arrS[i].ascii
            if num >= 48 && num <= 57
            {
                N /= num - 48
                K %= N
            }
            else if K % N == 0
            {
                return String(arrS[i])
            }
            else
            {
                N -= 1
            }
            i -= 1
        }
        return String()
    }
}

//Character扩展
extension Character  
{  
  //Character转ASCII整数值(定义小写为整数值)
   var ascii: Int {
       get {
           return Int(self.unicodeScalars.first?.value ?? 0)
       }       
    }
}

---

4ms

class Solution {
    func decodeAtIndex(_ S: String, _ K: Int) -> String {
        var stack: [Character] = []
        
        var charCount: Int = 0 
        var movingIndex: Int = 0 
        var S = Array(S)
        
        while movingIndex < S.count && charCount < K {
            let character = S[movingIndex]
            if character.isDigit() {
                charCount *= character.toDigit() 
            } else {
                charCount += 1
            }
            movingIndex += 1
        }
        var k = K 
        while movingIndex > 0 {
            movingIndex -= 1 
            let character = S[movingIndex]
            if character.isDigit() {
                charCount /= character.toDigit() 
                k %= charCount 
            } else {
                if k == 0 || k == charCount {
                    return String(character)
                }
                charCount -= 1 
            }
        }
        
        return "" 
    }
}

extension Character {
    func isDigit() -> Bool {
        return Int(String(self)) != nil 
    }
    
    func toDigit() -> Int {
        return Int(String(self))!
    }
}

---

12ms

class Solution {
    let digiSets = Set<Character>("123456789")
    func decodeAtIndex(_ S: String, _ K: Int) -> String {
        if K == 0 {
            if !digiSets.contains(S.last!) { return String(S.last!) }
            return decodeAtIndex(String(S.dropLast()), K)
        }

        var lastStr = [Character]()
        var lastCount = 0, curCount = 0
        var chars = Array(S)
        for i in chars.indices {
            if digiSets.contains(chars[i]) {
                curCount *= Int(String(chars[i]))!
                if curCount >= K { return decodeAtIndex(String(lastStr), K % lastCount)}
            }
            else {
                curCount += 1
                if curCount == K  { return String(chars[i]) }
            }
            lastStr.append(chars[i])
            lastCount = curCount
        }
        return ""
    }
}

---

20ms

class Solution {
    func decodeAtIndex(_ S: String, _ K: Int) -> String {        
        func isNum(_ char: Character) -> Bool {
            if char == "2" || char == "3" || char == "4" || char == "5" || char == "6" || char == "7" || char == "8" || char == "9" {
                return true
            }
            return false
        }
        var size = 0, k = K
        for char in S {
            if isNum(char) {
                let str = String(char)
                size *= (Int(str) ?? 1-1)
            } else {
                size += 1
            }
        }
        for item in S.reversed() {
            k %= size
            if k == 0  && !isNum(item) {
                return String(item)
            }
            if isNum(item) {
                let str = String(item)
                size /= (Int(str) ?? 1)
            } else {
                size -= 1
            }
        }
        return ""        
    }
}