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.

Analysis:
这道题思路很简单，但是Corner case很多。
1. 如果res 的size > 2, 当前取的val比target小也符合，因为后面会加更多的数进来。一旦大了就没必要继续了
2. 如果取得数大于1位，而且从第一位是0也不符合，直接break
3. 当前val大于Integer.MAX_VALUE
Solution:

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