[Swift]LeetCode306. 累加数 | Additive Number

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Additive number is a string whose digits can form additive sequence.

A valid additive sequence should contain at least three numbers. Except for the first two numbers, each subsequent number in the sequence must be the sum of the preceding two.

Given a string containing only digits '0'-'9', write a function to determine if it's an additive number.

Note: Numbers in the additive sequence cannot have leading zeros, so sequence 1, 2, 03 or 1, 02, 3 is invalid.

Example 1:

Input: "112358"
Output: true 
Explanation: The digits can form an additive sequence: 1, 1, 2, 3, 5, 8. 
             1 + 1 = 2, 1 + 2 = 3, 2 + 3 = 5, 3 + 5 = 8

Example 2:

Input: "199100199"
Output: true 
Explanation: The additive sequence is: 1, 99, 100, 199. 
             1 + 99 = 100, 99 + 100 = 199

Follow up:
How would you handle overflow for very large input integers?

---

累加数是一个字符串，组成它的数字可以形成累加序列。

一个有效的累加序列必须至少包含 3 个数。除了最开始的两个数以外，字符串中的其他数都等于它之前两个数相加的和。

给定一个只包含数字 '0'-'9' 的字符串，编写一个算法来判断给定输入是否是累加数。

说明: 累加序列里的数不会以 0 开头，所以不会出现 1, 2, 03 或者 1, 02, 3 的情况。

示例 1:

输入: "112358"
输出: true 
解释: 累加序列为: 1, 1, 2, 3, 5, 8 。1 + 1 = 2, 1 + 2 = 3, 2 + 3 = 5, 3 + 5 = 8

示例 2:

输入: "199100199"
输出: true 
解释: 累加序列为: 1, 99, 100, 199。1 + 99 = 100, 99 + 100 = 199

进阶:
你如何处理一个溢出的过大的整数输入?

---

16ms

class Solution {
    func isAdditiveNumber(_ num: String) -> Bool {
        let array = [Character](num)
        let n = array.count
        if n < 3 { return false }
        
        func helper(s1: String, s2: String, remain: String) -> Bool {
            let one = Int(s1)!
            let two = Int(s2)!
            let next = "(one + two)"
            if next == remain { return true }
            var newRemain = remain
            if remain.hasPrefix(next) {
                let range = remain.range(of: next)!
                newRemain.replaceSubrange(range, with: "")
                return helper(s1: s2, s2: next, remain: newRemain)
            } else {
                return false
            }
        }
        
        //设第一个数是以i结尾的，第二个数是以j结尾的
        for i in 0 ... (n - 1)/2 - 1 { //i只能取到一半以下的值
            let one = String(array[0...i])
            if one != "0" && one.hasPrefix("0") { continue }
            for j in i + 1 ... n - 2 - i { //j至少取一位，且 n - j >= i
                let two = String(array[i+1 ... j])
                if two != "0" && two.hasPrefix("0") { continue }
                let remain = String(array[j+1 ... n-1])
                if remain != "0" && remain.hasPrefix("0") { continue }
                if helper(s1: one, s2: two, remain: remain) {
                    return true
                }
            }
        }
        return false
    }
}

---

24ms

class Solution {
    func isAdditiveNumber(_ num: String) -> Bool {
        
        if num.count < 3 {
            return false
        }
        
        let numArr = Array(num)
        
        for i in 1..<numArr.count-1 {
            for j in i+1..<numArr.count {
                let f = String(numArr[0..<i])
                let s = String(numArr[i..<j])
                var fn = Int(f)!
                var sn = Int(s)!
                var ln = fn + sn
                
                var l = "(ln)"
                var total = f + s + l
                
                if (f.count > 1 && f.first == "0" )||(s.count > 1 && s.first == "0") {
                    continue
                }
                while total.count < num.count {
                    let totalArr = Array(total)
                    if totalArr != Array(numArr[0..<totalArr.count]) {
                        break
                    }
                    fn = sn
                    sn = ln
                    ln = fn + sn
                    l = "(ln)"
                    total += l
                }
                if total == num {
                    return true
                }
            }
        }
        return false
    }
}