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 all0 <= 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.
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 []。
- 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。