# [842. Split Array into Fibonacci Sequence](https://leetcode.com/problems/split-array-into-fibonacci-sequence/)


## 题目

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


## 题目大意

给定一个数字字符串 S，比如 S = "123456579"，我们可以将它分成斐波那契式的序列 [123, 456, 579]。斐波那契式序列是一个非负整数列表 F，且满足：

- 0 <= F[i] <= 2^31 - 1，（也就是说，每个整数都符合 32 位有符号整数类型）；
- F.length >= 3；
- 对于所有的0 <= i < F.length - 2，都有 F[i] + F[i+1] = F[i+2] 成立。

另外，请注意，将字符串拆分成小块时，每个块的数字一定不要以零开头，除非这个块是数字 0 本身。返回从 S 拆分出来的所有斐波那契式的序列块，如果不能拆分则返回 []。



## 解题思路


- 这一题是第 306 题的加强版。第 306 题要求判断字符串是否满足斐波那契数列形式。这一题要求输出按照斐波那契数列形式分割之后的数字数组。
- 这一题思路和第 306 题基本一致，需要注意的是题目中的一个限制条件，`0 <= F[i] <= 2^31 - 1`，注意这个条件，笔者开始没注意，后面输出解就出现错误了，可以看笔者的测试文件用例的最后两组数据，这两组都是可以分解成斐波那契数列的，但是由于分割以后的数字都大于了 `2^31 - 1`，所以这些解都不能要！
- 这一题也要特别注意剪枝条件，没有剪枝条件，时间复杂度特别高，加上合理的剪枝条件以后，0ms 通过。



## 代码

```go

package leetcode

import (
	"strconv"
	"strings"
)

func splitIntoFibonacci(S string) []int {
	if len(S) < 3 {
		return []int{}
	}
	res, isComplete := []int{}, false
	for firstEnd := 0; firstEnd < len(S)/2; firstEnd++ {
		if S[0] == '0' && firstEnd > 0 {
			break
		}
		first, _ := strconv.Atoi(S[:firstEnd+1])
		if first >= 1<<31 { // 题目要求每个数都要小于 2^31 - 1 = 2147483647，此处剪枝很关键！
			break
		}
		for secondEnd := firstEnd + 1; max(firstEnd, secondEnd-firstEnd) <= len(S)-secondEnd; secondEnd++ {
			if S[firstEnd+1] == '0' && secondEnd-firstEnd > 1 {
				break
			}
			second, _ := strconv.Atoi(S[firstEnd+1 : secondEnd+1])
			if second >= 1<<31 { // 题目要求每个数都要小于 2^31 - 1 = 2147483647，此处剪枝很关键！
				break
			}
			findRecursiveCheck(S, first, second, secondEnd+1, &res, &isComplete)
		}
	}
	return res
}

//Propagate for rest of the string
func findRecursiveCheck(S string, x1 int, x2 int, left int, res *[]int, isComplete *bool) {
	if x1 >= 1<<31 || x2 >= 1<<31 { // 题目要求每个数都要小于 2^31 - 1 = 2147483647，此处剪枝很关键！
		return
	}
	if left == len(S) {
		if !*isComplete {
			*isComplete = true
			*res = append(*res, x1)
			*res = append(*res, x2)
		}
		return
	}
	if strings.HasPrefix(S[left:], strconv.Itoa(x1+x2)) && !*isComplete {
		*res = append(*res, x1)
		findRecursiveCheck(S, x2, x1+x2, left+len(strconv.Itoa(x1+x2)), res, isComplete)
		return
	}
	if len(*res) > 0 && !*isComplete {
		*res = (*res)[:len(*res)-1]
	}
	return
}

```