title
A sequence X_1, X_2, …, X_n is fibonacci-like if:
n >= 3 X_i + X_{i+1} = X_{i+2} for all i + 2 <= n Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, without changing the order of the remaining elements. For example, [3, 5, 8] is a subsequence of [3, 4, 5, 6, 7, 8].)
Example 1:
Input: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [1,3,7,11,12,14,18]
Output: 3
Explanation:
The longest subsequence that is fibonacci-like:
[1,11,12], [3,11,14] or [7,11,18].
solution
i, j , k是个fibonacci-like,
i用map存储，查找，查找后i< j 则合法
用一个map存当前结尾的最大fibonacci-like长度，
easy dp
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
index = {A[i]: i for i in range(len(A))}
longest = {}
ans = 0
length = len(A)
for i in range(length):
for j in range(i):
x = A[i] - A[j]
m = index.get(x)
if m is not None and m < j:
longest[(j, i)] = longest.get((m, j)) + 1 if longest.get((m, j)) is not None else 3
ans = max(ans, longest[(j, i)])
else:
continue
return ans if ans >= 3 else 0
class Solution {
public:
int lenLongestFibSubseq(vector<int>& A) {
int N = A.size();
map<int, int> index, longest;
for (int i = 0; i < N; i++) {
index[A[i]] = i;
}
int ans = 0;
for (int k = 0; k < N; k++) {
for (int j = 0; j < k; j++) {
if (index.count(A[k]-A[j])) {
int i = index[A[k] - A[j]];
if (i < j) {
if (longest.count(i*N + j)) {
longest[j*N + k] = longest[i*N + j]+1;
}
else {
longest[j*N + k] = 3;
}
ans = max(ans, longest[j*N + k]);
}
}
}
}
return ans >= 3 ? ans : 0;
}
};