Leetcode之Additive Number

題目：

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

Constraints:

• num consists only of digits '0'-'9'.
• 1 <= num.length <= 35

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

代碼：

方法一——迭代法：

class Solution {
public:
    bool isAdditiveNumber(string num) {
        for (int i = 1; i < num.size(); ++i) {
            string s1 = num.substr(0, i);
            if (s1.size() > 1 && s1[0] == '0') break;
            for (int j = i + 1; j < num.size(); ++j) {
                string s2 = num.substr(i, j - i);
                long d1 = stol(s1), d2 = stol(s2);
                if ((s2.size() > 1 && s2[0] == '0')) break;
                long next = d1 + d2;
                string nextStr = to_string(next);
                if (nextStr != num.substr(j, nextStr.length())) continue;
                string allStr = s1 + s2 + nextStr;
                while (allStr.size() < num.size()) {
                    d1 = d2;
                    d2 = next;
                    next = d1 + d2;
                    nextStr = to_string(next);
                    allStr += nextStr;
                }
                if (allStr == num) return true;
            }
        }
        return false;
    }
};

方法二——遞歸法：

class Solution {
public:
    bool isAdditiveNumber(string num) {
        bool res = false;
        vector<long> out;
        helper(num, 0, out, res);
        return res;
    }
    void helper(string& num, int start, vector<long>& out, bool& res) {
        if (res) return;
        if (start >= num.size()) {
            if (out.size() >= 3) res = true; 
            return;
        }
        for (int i = start; i < num.size(); ++i) {
            string str = num.substr(start, i - start + 1);
            if ((str.size() > 1 && str[0] == '0') || str.size() > 19) break;
            long curNum = stol(str), n = out.size();
            if (out.size() >= 2 && curNum != out[n - 1] + out[n - 2]) continue;
            out.push_back(curNum);
            helper(num, i + 1, out, res);
            out.pop_back();
        }
    }
};