[Swift]LeetCode906. 超级回文数 | Super Palindromes

★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤微信公众号：山青咏芝（shanqingyongzhi）
➤博客园地址：山青咏芝（https://www.cnblogs.com/strengthen/）
➤GitHub地址：https://github.com/strengthen/LeetCode
➤原文地址： https://www.cnblogs.com/strengthen/p/10609103.html 
➤如果链接不是山青咏芝的博客园地址，则可能是爬取作者的文章。
➤原文已修改更新！强烈建议点击原文地址阅读！支持作者！支持原创！
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★

Let's say a positive integer is a superpalindrome if it is a palindrome, and it is also the square of a palindrome.

Now, given two positive integers L and R (represented as strings), return the number of superpalindromes in the inclusive range [L, R]. 

Example 1:

Input: L = "4", R = "1000"
Output: 4
Explanation: 4, 9, 121, and 484 are superpalindromes.
Note that 676 is not a superpalindrome: 26 * 26 = 676, but 26 is not a palindrome. 

Note:

1. 1 <= len(L) <= 18
2. 1 <= len(R) <= 18
3. L and R are strings representing integers in the range [1, 10^18).
4. int(L) <= int(R)

---

如果一个正整数自身是回文数，而且它也是一个回文数的平方，那么我们称这个数为超级回文数。

现在，给定两个正整数 L 和 R （以字符串形式表示），返回包含在范围 [L, R] 中的超级回文数的数目。 

示例：

输入：L = "4", R = "1000"
输出：4
解释：
4，9，121，以及 484 是超级回文数。
注意 676 不是一个超级回文数： 26 * 26 = 676，但是 26 不是回文数。 

提示：

1. 1 <= len(L) <= 18
2. 1 <= len(R) <= 18
3. L 和 R 是表示 [1, 10^18) 范围的整数的字符串。
4. int(L) <= int(R)

---

Runtime: 8 ms

Memory Usage: 19.9 MB

class Solution {
    let value:[Int] = [0, 1, 4, 9, 121, 484, 676, 10201, 12321, 14641, 40804, 44944, 69696, 94249, 698896, 1002001, 1234321, 
    4008004, 5221225, 6948496, 100020001, 102030201, 104060401, 121242121, 123454321, 125686521, 400080004, 
    404090404, 522808225, 617323716, 942060249, 10000200001, 10221412201, 12102420121, 12345654321, 
    40000800004, 637832238736, 1000002000001, 1002003002001, 1004006004001, 1020304030201, 1022325232201, 
    1024348434201, 1086078706801, 1210024200121, 1212225222121, 1214428244121, 1230127210321, 1232346432321, 
    1234567654321, 1615108015161, 4000008000004, 4004009004004, 4051154511504, 5265533355625, 9420645460249, 
    100000020000001, 100220141022001, 102012040210201, 102234363432201, 121000242000121, 121242363242121, 
    123212464212321, 123456787654321, 123862676268321, 144678292876441, 165551171155561, 400000080000004, 
    900075181570009, 4099923883299904, 10000000200000001, 10002000300020001, 10004000600040001, 10020210401202001, 
    10022212521222001, 10024214841242001, 10201020402010201, 10203040504030201, 10205060806050201, 
    10221432623412201, 10223454745432201, 12100002420000121, 12102202520220121, 12104402820440121, 
    12120030703002121, 12122232623222121, 12124434743442121, 12321024642012321, 12323244744232321, 
    12341234943214321, 12343456865434321, 12345678987654321, 40000000800000004, 40004000900040004, 94206450305460249]
    func superpalindromesInRange(_ L: String, _ R: String) -> Int {
        var l:Int = bound(Int(L) ?? 0)
        var r:Int = bound(Int(R) ?? 0)
        var res:Int = 0
        while(l != r)
        {
            var v:Int = value[l]
            var root:Int = Int(sqrt(Double(v)))
            var s1:String = String(root)
            var s2:String = String(s1.reversed())
            if s2 == s1
            {
                res += 1
            }
            l += 1
        }
        return res        
    }

    func bound(_ target:Int) -> Int
    {
        var low = 0
        var high = value.count - 1
        var mid = (low + high) >> 1
        
        while low <= high {
            let val = value[mid]
            if target == val {
                return mid
            } else if target < val {
                high = mid - 1
            } else {
                low = mid + 1 
            }
            mid = (low + high) >> 1
        }
        return low
    }
}

---

Time Limit Exceeded

class Solution {
    func superpalindromesInRange(_ L: String, _ R: String) -> Int {
        var l:Int = Int(L) ?? 0
        var r:Int = Int(R) ?? 0
        var result:Int = 0
        var i:Int = Int(sqrt(Double(l)))
        while(i * i <= r)
        {
            var p:Int = nextP(i)
            if p * p <= r && isP(p * p)
            {
                result += 1
            }
            i = p + 1
        }
        return result
    }

    func nextP(_ l:Int) -> Int
    {
        var s:String = String(l)
        var len:Int = s.count
        var cands:[Int] = [Int]()
        let num:Int = Int(pow(10, Double(len))) - 1
        cands.append(num)
        var half:String = s.subString(0, (len + 1) / 2)
        let num0:Int? = Int(half)
        var nextHalf:String = String(1)
        if num0 != nil
        {
            nextHalf = String(num0! + 1)
        }
        var reverse:String = String(half.subString(0, len / 2).reversed())
        var nextReverse:String = String(nextHalf.subString(0, len / 2).reversed())
        let num1:Int? = Int(half + reverse)
        if num1 !=  nil
        {
            cands.append(num1!)
        }
        let num2:Int? = Int(nextHalf + nextReverse)
        if num2 !=  nil
        {
            cands.append(num2!)
        }
        var result:Int = Int.max
        for i in cands
        {
            if i >= l
            {
                result = min(result, i)
            }
        }
        return result
    }

    func isP(_ l:Int) -> Bool
    {
        var arrS:[Character] = Array(String(l))
        var i:Int = 0
        var j:Int = arrS.count - 1
        while (i < j)
        {
            if arrS[i] != arrS[j]
            {
                return false
            }       
            i += 1
            j -= 1     
        }
        return true
    }
}

extension String {
    // 截取字符串：指定索引和字符数
    // - begin: 开始截取处索引
    // - count: 截取的字符数量
    func subString(_ begin:Int,_ count:Int) -> String {
        let start = self.index(self.startIndex, offsetBy: max(0, begin))
        let end = self.index(self.startIndex, offsetBy:  min(self.count, begin + count))
        return String(self[start..<end]) 
    }
}