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 <= S.length <= 200
S contains only digits.
这题和Additive Number一样的,但是需要注意的是溢出后就可以不继续计算下去了。
class SplitArrayIntoFibonacciSequence : public Solution {
public:
void Exec() {
}
int64_t str2int(string &num, int at, int len) {
int64_t value = 0;
int64_t base = 1;
char lastChar = 0;
for (int i = at + len - 1; i >= at; i--) {
value += (num[i] - '0') * base;
base *= 10;
lastChar = num[i];
}
if (lastChar == '0' && len > 1) return -1;
return value;
}
vector<int> splitIntoFibonacci(string S) {
if (S.size() <= 2) return vector<int>();
vector<int> res;
for (int flen = 1; flen <= std::min(int(S.size()) / 2, 10); flen++) {
for (int slen = 1; slen <= std::min(int(S.size()) / 2, 10); slen++) {
res.clear();
int index = flen + slen;
int64_t fv = str2int(S, 0, flen);
int64_t sv = str2int(S, flen, slen);
if (fv < 0 || sv < 0) continue;
res.push_back(fv);
res.push_back(sv);
int nexti = flen + slen;
string want = std::to_string(fv + sv);
while (S.size() - 1 - nexti + 1 >= want.size() &&
0 == memcmp(S.c_str() + nexti, want.c_str(),
want.size()) &&
fv + sv <= INT_MAX) {
res.push_back(fv + sv);
nexti += want.size();
if (nexti >= S.size()) return res;
int64_t nextv = fv + sv;
fv = sv;
sv = nextv;
want = std::to_string(fv + sv);
}
}
}
res.clear();
return res;
}
};