873 Length of Longest Fibonacci Subsequence 最長的斐波那契子序列的長度
Description:
A sequence x1, x2, ..., xn is Fibonacci-like if:
n >= 3
xi + xi+1 == xi+2 for all i + 2 <= n
Given a strictly increasing array arr of positive integers forming a sequence, return the length of the longest Fibonacci-like subsequence of arr. If one does not exist, return 0.
A subsequence is derived from another sequence arr by deleting any number of elements (including none) from arr, without changing the order of the remaining elements. For example, [3, 5, 8] is a subsequence of [3, 4, 5, 6, 7, 8].
Example:
Example 1:
Input: arr = [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: arr = [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].
Constraints:
3 <= arr.length <= 1000
1 <= arr[i] < arr[i + 1] <= 10^9
題目描述:
如果序列 X_1, X_2, ..., X_n 滿足下列條件赏僧，就說它是 斐波那契式 的：
n >= 3
對于所有 i + 2 <= n玩讳，都有 X_i + X_{i+1} = X_{i+2}
給定一個嚴格遞增的正整數(shù)數(shù)組形成序列 arr 篮幢，找到 arr 中最長的斐波那契式的子序列的長度下面。如果一個不存在唠亚，返回 0 帝际。
（回想一下餐禁，子序列是從原序列 arr 中派生出來的诵肛，它從 arr 中刪掉任意數(shù)量的元素（也可以不刪）镀虐，而不改變其余元素的順序箱蟆。例如， [3, 5, 8] 是 [3, 4, 5, 6, 7, 8] 的一個子序列）
示例 :
示例 1：
輸入: arr = [1,2,3,4,5,6,7,8]
輸出: 5
解釋: 最長的斐波那契式子序列為 [1,2,3,5,8] 刮便。
示例 2：
輸入: arr = [1,3,7,11,12,14,18]
輸出: 3
解釋: 最長的斐波那契式子序列有 [1,11,12]空猜、[3,11,14] 以及 [7,11,18] 。
提示:
3 <= arr.length <= 1000
1 <= arr[i] < arr[i + 1] <= 10^9
思路:
動態(tài)規(guī)劃
dp 記錄以 arr[i], arr[j] 結(jié)尾的最長斐波拉契子數(shù)列的長度
m 記錄下標對應(yīng)的值, 方便快速查找
如果油用完了還不能到達 target 返回 -1
時間復(fù)雜度為 O(n ^ 2), 空間復(fù)雜度為 O(nlgm), 其中 m = max(arr)
代碼:
C++:
class Solution
{
public:
int lenLongestFibSubseq(vector<int>& arr)
{
int n = arr.size(), result = 0;
unordered_map<int, int> m, dp;
for (int i = 0; i < n; i++) m[arr[i]] = i;
for (int i = 0; i < n; i++)
{
for (int j = 0; j < i; j++)
{
if (arr[i] - arr[j] < arr[j] and m.count(arr[i] - arr[j]))
{
int k = m[arr[i] - arr[j]];
dp[j * n + i] = dp[k * n + j] + 1;
result = max(result, dp[j * n + i] + 2);
}
}
}
return result > 2 ? result : 0;
}
};
Java:
class Solution {
public int lenLongestFibSubseq(int[] arr) {
int n = arr.length, result = 0;
Map<Integer, Integer> m = new HashMap<>(), dp = new HashMap<>();
for (int i = 0; i < n; i++) m.put(arr[i], i);
for (int i = 0; i < n; i++) {
for (int j = 0; j < i; j++) {
if (arr[i] - arr[j] < arr[j] && m.containsKey(arr[i] - arr[j])) {
int k = m.get(arr[i] - arr[j]);
dp.put(j * n + i, dp.getOrDefault(k * n + j, 0) + 1);
result = Math.max(result, dp.getOrDefault(j * n + i, 0) + 2);
}
}
}
return result > 2 ? result : 0;
}
}
Python:
class Solution:
def lenLongestFibSubseq(self, arr: List[int]) -> int:
n, result, m, dp = len(arr), 0, { v: i for i, v in enumerate(arr) }, defaultdict(int)
for i in range(n):
for j in range(i):
if (cur := arr[i] - arr[j]) < arr[j] and cur in m:
dp[j * n + i], result = dp[m[cur] * n + j] + 1, max(result, dp[m[cur] * n + j] + 3)
return result if result > 2 else 0
