[Swift]LeetCode848. 字母移位 | Shifting Letters

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We have a string S of lowercase letters, and an integer array shifts.

Call the shift of a letter, the next letter in the alphabet, (wrapping around so that 'z' becomes 'a'). 

For example, shift('a') = 'b', shift('t') = 'u', and shift('z') = 'a'.

Now for each shifts[i] = x, we want to shift the first i+1 letters of S, x times.

Return the final string after all such shifts to S are applied.

Example 1:

Input: S = "abc", shifts = [3,5,9]
Output: "rpl"
Explanation: 
We start with "abc".
After shifting the first 1 letters of S by 3, we have "dbc".
After shifting the first 2 letters of S by 5, we have "igc".
After shifting the first 3 letters of S by 9, we have "rpl", the answer.

Note:

1. 1 <= S.length = shifts.length <= 20000
2. 0 <= shifts[i] <= 10 ^ 9

---

有一个由小写字母组成的字符串 S，和一个整数数组 shifts。

我们将字母表中的下一个字母称为原字母的 移位（由于字母表是环绕的， 'z' 将会变成 'a'）。

例如·，shift('a') = 'b'， shift('t') = 'u',， 以及 shift('z') = 'a'。

对于每个 shifts[i] = x ， 我们会将 S 中的前 i+1 个字母移位 x 次。

返回将所有这些移位都应用到 S 后最终得到的字符串。

示例：

输入：S = "abc", shifts = [3,5,9]
输出："rpl"
解释： 
我们以 "abc" 开始。
将 S 中的第 1 个字母移位 3 次后，我们得到 "dbc"。
再将 S 中的前 2 个字母移位 5 次后，我们得到 "igc"。
最后将 S 中的这 3 个字母移位 9 次后，我们得到答案 "rpl"。

提示：

1. 1 <= S.length = shifts.length <= 20000
2. 0 <= shifts[i] <= 10 ^ 9

---

Runtime: 340 ms

Memory Usage: 21 MB

class Solution {
    func shiftingLetters(_ S: String, _ shifts: [Int]) -> String {
        var arrS = Array(S)
        var i:Int = shifts.count - 1
        var m:Int = 0
        while(i >= 0)
        {
            m += shifts[i]
            arrS[i] = (((arrS[i].ascii - 97) + m) % 26 + 97).ASCII   
            i -= 1
            m %= 26            
        }
        return String(arrS)
    }
}

//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)
       }       
    }
}

//Int扩展
extension Int
{
    //Int转Character,ASCII值(定义大写为字符值)
    var ASCII:Character 
    {
        get {return Character(UnicodeScalar(self)!)}
    }
}

---

652ms

class Solution {
    func shiftingLetters(_ S: String, _ shifts: [Int]) -> String {
    var number:UInt32 = 0
        var i = 0;
        var count = 0 ;
        var string : String = ""
        let startCount :UInt32 = 97
        
        var newShifts :[Int] = []  //换算出一个新的数组
        var allCount = 0  //算出总数
        for val in shifts{
            let newval = val%26
            newShifts .append(newval)
            allCount += newval
        }
        for code in S.unicodeScalars {
            number = code.value - startCount
            
            number = (number +  UInt32(allCount))%26
            var ch:Character = Character(UnicodeScalar(number+startCount)!)
            string.append(ch);
            
            if(i < newShifts.count)
            {
                allCount -= newShifts[i]
            }
            else{
                allCount = 0
            }
            i += 1
        }
        return string
    }
}

---

1568ms

class Solution {
        func shiftingLetters(_ S: String, _ shifts: [Int]) -> String {
        
        func move(s:String,steps:[Int]) -> String{
            let lastChar = Int("z".unicodeScalars.first!.value)
            let firstChar = Int("a".unicodeScalars.first!.value)
            let lenth = lastChar - firstChar + 1
            let chars = zip(s,steps).map { (arg) -> Character in
                print(arg.0,arg.0.unicodeScalars.first!.value)
                let value = ((Int(arg.0.unicodeScalars.first!.value) - firstChar) + arg.1)%lenth + firstChar
                
                return Character.init(Unicode.Scalar.init(value)!)
                
                
                
            }
            return String.init(chars)
        }
        
        var steps:[Int] = []
       
        var lastInt = 0
        for i in stride(from: shifts.count-1, through: 0, by: -1) {
            lastInt += shifts[i]
            steps.append(lastInt)
        }
        return move(s:S ,steps:steps.reversed())
    }
}