[LeetCode] 842. Split Array into Fibonacci Sequence

Problem

Given a string S of digits, such as S = "123456579", we can split it into a Fibonacci-like sequence [123, 456, 579].

Formally, a Fibonacci-like sequence is a list F of non-negative integers such that:

0 <= F[i] <= 2^31 - 1, (that is, each integer fits a 32-bit signed integer type);
F.length >= 3;
and F[i] + F[i+1] = F[i+2] for all 0 <= i < F.length - 2.
Also, note that when splitting the string into pieces, each piece must not have extra leading zeroes, except if the piece is the number 0 itself.

Return any Fibonacci-like sequence split from S, or return [] if it cannot be done.

Example 1:

Input: "123456579"
Output: [123,456,579]
Example 2:

Input: "11235813"
Output: [1,1,2,3,5,8,13]
Example 3:

Input: "112358130"
Output: []
Explanation: The task is impossible.
Example 4:

Input: "0123"
Output: []
Explanation: Leading zeroes are not allowed, so "01", "2", "3" is not valid.
Example 5:

Input: "1101111"
Output: [110, 1, 111]
Explanation: The output [11, 0, 11, 11] would also be accepted.
Note:

1 <= S.length <= 200
S contains only digits.

Solution

class Solution {
    public List<Integer> splitIntoFibonacci(String S) {
        List<Integer> res = new ArrayList<>();
        if (S == null) return res;
        helper(S, 0, res);
        return res;
    }
    private boolean helper(String str, int start, List<Integer> res) {
        if (start == str.length() && res.size() > 2) return true;
        for (int i = start; i < str.length(); i++) {
            if (str.charAt(start) == '0' && i != start) break;
            long num = Long.parseLong(str.substring(start, i+1));
            if (num > Integer.MAX_VALUE) return false;
            int size = res.size();
            if (size >= 2 && num > res.get(size-2) + res.get(size-1)) break;
            if (size <= 1 || num == res.get(size-2) + res.get(size-1)) {
                res.add((int)num);
                if (helper(str, i+1, res)) return true;
                res.remove(res.size()-1);
            }
        }
        return false;
    }
}