[Swift]LeetCode400. 第N个数字 | Nth Digit

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Find the nth digit of the infinite integer sequence 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, ...

Note:
n is positive and will fit within the range of a 32-bit signed integer (n < 231).

Example 1:

Input:
3

Output:
3

Example 2:

Input:
11

Output:
0

Explanation:
The 11th digit of the sequence 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, ... is a 0, which is part of the number 10.

---

在无限的整数序列 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, ...中找到第 n 个数字。

注意:
n 是正数且在32为整形范围内 ( n < 231)。

示例 1:

输入:
3

输出:
3

示例 2:

输入:
11

输出:
0

说明:
第11个数字在序列 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, ... 里是0，它是10的一部分。

---

8ms

class Solution {
    func findNthDigit(_ n: Int) -> Int {
        var len:Int = 1
        var base:Int = 1
        var num:Int = n
        while(n > 9*base*len)
        {
            num -= 9*base*len
            len += 1
            base *= 10
        }
        var curNum:Int = (num-1)/len + base
        var digit:Int = 0
        var start:Int = (num-1)%len 
        for i in start..<len
        {
            digit = curNum%10
            curNum /= 10
        }
        return digit
    }
}

---

8ms

class Solution {
    func findNthDigit(_ n: Int) -> Int {
        guard n >= 10 else {
            return n
        }
        
        var index = 9
        var a = 90
        var zero = 2
        var aa = 9
        
        while index + a * zero < n {
            index += a * zero
            zero += 1
            a *= 10
            aa = aa * 10 + 9
        }

        let number = Int(String(Array(String(d))[e]))!
        return number
    }
}

---

8ms

class Solution {
    func findNthDigit(_ n: Int) -> Int {
        if n <= 0 { return 0 }
        
        // Find the interval in which the digit will be
        var digitMultiplier = 1 // number of digits of numbers in current interval
        var fromLimit = 1 // start of the interval
        var toLimit = 10 // end of the interval
        var passedDigitCount = 0 // count of digits of all numbers in passed intervals
        var nextLimitPassedDigitCount = 9 // number of passed digits if we move to the next interval
        
        while n > nextLimitPassedDigitCount {
            passedDigitCount = nextLimitPassedDigitCount
            fromLimit = toLimit
            toLimit *= 10
            digitMultiplier += 1
            nextLimitPassedDigitCount += (toLimit - fromLimit) * digitMultiplier
        }
        
        let n = n - passedDigitCount
        
        var index = n % digitMultiplier;
        var number = 0
        if index == 0 {
            index = digitMultiplier
            number = fromLimit + n / digitMultiplier - 1
        } else {
            number = fromLimit + n / digitMultiplier 
        }
               
        for _ in index..<digitMultiplier {
            number /= 10
        }
        
        return number % 10;
               
    }
}

---

12ms

class Solution {
    func findNthDigit(_ n: Int) -> Int {
        var ar: [Int] = [9, 180, 2700]
        var length = 1, multiplier = 9, startNumber = 1
        var nth = n
        while nth - length * multiplier > 0 {
            nth -= length * multiplier
            length += 1
            multiplier *= 10
            startNumber *= 10 
        }
        
        var numberAtNth = startNumber + (nth-1)/length
        var str = String(numberAtNth)
        let index = str.index(str.startIndex, offsetBy: (nth-1)%length)
        return Int(String(str[index]))!
    }
}

---

16ms

class Solution {
    func f(_ n: Int) -> Int {
        var idx = 0 // 1 base
        var block_idx = 0 // 1 base
        var item_size = 0 // = block_idx
        var item_cnt = 0
        var block_size = 0
        repeat
        {
            block_idx += 1
            item_size = block_idx
            if item_cnt == 0 {
                item_cnt = 9
            } else {
                item_cnt *= 10
            }
            block_size = item_cnt * item_size
            idx += block_size
        } while idx < n
        var in_block_idx = n - (idx - block_size) - 1 // 0 base
        var item_idx = in_block_idx / item_size // 0 base
        var in_item_idx = in_block_idx % item_size // 0 base
        
        var first_item = 1;
        if (block_idx - 1 > 0) {
            for i in 1...block_idx - 1 {
                first_item *= 10
            }
        }
        var item = first_item + item_idx
        var tgt = item
        var do_times = item_size - in_item_idx - 1
        if do_times > 0 {
            for i in 1...do_times {
                tgt /= 10
            }
        }
        tgt %= 10
        return tgt        
    }
    
    func findNthDigit(_ n: Int) -> Int {
        return f(n)
    }
}