[Swift]LeetCode386. 字典序排数 | Lexicographical Numbers

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Given an integer n, return 1 - n in lexicographical order.

For example, given 13, return: [1,10,11,12,13,2,3,4,5,6,7,8,9].

Please optimize your algorithm to use less time and space. The input size may be as large as 5,000,000.

---

给定一个整数 n, 返回从 1 到 n 的字典顺序。

例如，

给定 n =1 3，返回 [1,10,11,12,13,2,3,4,5,6,7,8,9] 。

请尽可能的优化算法的时间复杂度和空间复杂度。 输入的数据 n 小于等于 5,000,000。

---

112ms

class Solution {
    func lexicalOrder(_ n: Int) -> [Int] {
        var res:[Int] = [Int]()
        for i in 1...9
        {
            helper(i, n, &res)
        }
        return res
    }
    
    func helper(_ cur:Int,_ n:Int,_ res:inout[Int])
    {
        if cur > n {return}
        res.append(cur)
        for i in 0...9
        {
            if cur * 10 + i <= n
            {
                helper(cur * 10 + i, n, &res)
            }
            else
            {
                break
            }
        }
    }
}

---

152ms

class Solution {
    func lexicalOrder(_ n: Int) -> [Int] {
        var res = [Int]()
        for i in 1 ... 9 {
            dfs(i, n, &res)
        }
        
        return res
    }
    
    func dfs(_ cur: Int, _ n: Int, _ res: inout [Int]) {
        if cur > n {
            return
        }
        res.append(cur)
        for i in 0 ... 9 {
            if cur * 10 + i > n {
                return
            }
            dfs(cur * 10 + i, n, &res)
        }
    }
}

---

184ms

class Solution {
    func lexicalOrder(_ n: Int) -> [Int] {
        var result = [Int]()
        var current = 1
        for i in 1 ... n {
            result.append(current)
            if current * 10 <= n {
                current = current * 10
            } else if current % 10 != 9 && current + 1 <= n {
                current += 1
            } else {
                while (current / 10) % 10 == 9 {
                    current = current / 10
                }
                current = current / 10 + 1
            }
        }
        return result
    }
}