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website/content/ChapterFour/0842.Split-Array-into-Fibonacci-Sequence.md

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+# [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
+}
+
+```