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剑指 Offer II 093. 最长斐波那契数列

题目描述

如果序列 X_1, X_2, ..., X_n 满足下列条件,就说它是 斐波那契式 的:

  • n >= 3
  • 对于所有 i + 2 <= n,都有 X_i + X_{i+1} = X_{i+2}

给定一个严格递增的正整数数组形成序列 arr ,找到 arr 中最长的斐波那契式的子序列的长度。如果一个不存在,返回  0 。

(回想一下,子序列是从原序列  arr 中派生出来的,它从 arr 中删掉任意数量的元素(也可以不删),而不改变其余元素的顺序。例如, [3, 5, 8] 是 [3, 4, 5, 6, 7, 8] 的一个子序列)

 

示例 1:

输入: arr = [1,2,3,4,5,6,7,8]
输出: 5
解释: 最长的斐波那契式子序列为 [1,2,3,5,8] 。

示例 2:

输入: arr = [1,3,7,11,12,14,18]
输出: 3
解释: 最长的斐波那契式子序列有 [1,11,12]、[3,11,14] 以及 [7,11,18] 。

 

提示:

  • 3 <= arr.length <= 1000
  • 1 <= arr[i] < arr[i + 1] <= 10^9

 

注意:本题与主站 873 题相同: https://leetcode.cn/problems/length-of-longest-fibonacci-subsequence/

解法

方法一

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class Solution:
    def lenLongestFibSubseq(self, arr: List[int]) -> int:
        mp = {v: i for i, v in enumerate(arr)}
        n = len(arr)
        dp = [[0] * n for _ in range(n)]
        for i in range(n):
            for j in range(i):
                dp[j][i] = 2
        ans = 0
        for i in range(n):
            for j in range(i):
                delta = arr[i] - arr[j]
                if delta in mp:
                    k = mp[delta]
                    if k < j:
                        dp[j][i] = dp[k][j] + 1
                        ans = max(ans, dp[j][i])
        return ans
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class Solution {
    public int lenLongestFibSubseq(int[] arr) {
        int n = arr.length;
        Map<Integer, Integer> mp = new HashMap<>();
        for (int i = 0; i < n; ++i) {
            mp.put(arr[i], i);
        }
        int[][] dp = new int[n][n];
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < i; ++j) {
                dp[j][i] = 2;
            }
        }
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < i; ++j) {
                int delta = arr[i] - arr[j];
                if (mp.containsKey(delta)) {
                    int k = mp.get(delta);
                    if (k < j) {
                        dp[j][i] = dp[k][j] + 1;
                        ans = Math.max(ans, dp[j][i]);
                    }
                }
            }
        }
        return ans;
    }
}
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class Solution {
public:
    int lenLongestFibSubseq(vector<int>& arr) {
        unordered_map<int, int> mp;
        int n = arr.size();
        for (int i = 0; i < n; ++i) mp[arr[i]] = i;
        vector<vector<int>> dp(n, vector<int>(n));
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < i; ++j)
                dp[j][i] = 2;
        }
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < i; ++j) {
                int delta = arr[i] - arr[j];
                if (mp.count(delta)) {
                    int k = mp[delta];
                    if (k < j) {
                        dp[j][i] = dp[k][j] + 1;
                        ans = max(ans, dp[j][i]);
                    }
                }
            }
        }
        return ans;
    }
};
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func lenLongestFibSubseq(arr []int) int {
    n := len(arr)
    mp := make(map[int]int, n)
    for i, v := range arr {
        mp[v] = i + 1
    }
    dp := make([][]int, n)
    for i := 0; i < n; i++ {
        dp[i] = make([]int, n)
        for j := 0; j < i; j++ {
            dp[j][i] = 2
        }
    }
    ans := 0
    for i := 0; i < n; i++ {
        for j := 0; j < i; j++ {
            delta := arr[i] - arr[j]
            k := mp[delta] - 1
            if k >= 0 && k < j {
                dp[j][i] = dp[k][j] + 1
                ans = max(ans, dp[j][i])
            }
        }
    }
    return ans
}
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class Solution {
    func lenLongestFibSubseq(_ arr: [Int]) -> Int {
        let n = arr.count
        var mp = [Int: Int]()
        for i in 0..<n {
            mp[arr[i]] = i
        }

        var dp = Array(repeating: Array(repeating: 2, count: n), count: n)
        var ans = 0

        for i in 0..<n {
            for j in 0..<i {
                let delta = arr[i] - arr[j]
                if let k = mp[delta], k < j {
                    dp[j][i] = dp[k][j] + 1
                    ans = max(ans, dp[j][i])
                }
            }
        }

        return ans > 2 ? ans : 0
    }
}

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