0842. Split Array Into Fibonacci Sequence

842. 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: []
``````

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 #

• 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。

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 #

``````
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
}

``````

Nov 25, 2022