873. Length of Longest Fibonacci Subsequence

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].

Note:

• 3 <= A.length <= 1000
• 1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
• (The time limit has been reduced by 50% for submissions in Java, C, and C++.)

Approach #1: unordered_map. [C++]

class Solution {
public:
    int lenLongestFibSubseq(vector<int>& A) {
        unordered_map<int, int> memo;
        int len = A.size();
        int ans = 0, temp = 0;
        for (int i = 0; i < len; ++i)
            memo[A[i]] = i;
        for (int i = 0; i < len; ++i) {
            for (int j = i + 1; j < len; ++j) {
                int ant = 2;
                int last_idx = i;
                for (int cur_idx = j; cur_idx < len; ) {
                    temp = A[last_idx] + A[cur_idx];
                    if (memo.count(temp)) {
                        ant++;
                        last_idx = cur_idx;
                        cur_idx = memo[temp];
                    } else break;
                }
                ans = max(ans, ant);
            }
        }
        return ans == 2 ? 0 : ans;
    }
};

Approach #2: DP. [Java]

class Solution {
    public int lenLongestFibSubseq(int[] A) {
        int n = A.length;
        int res = 0;
        int[][] dp = new int[n+1][n+1];
        for (int[] row : dp) Arrays.fill(row, 2);
        Map<Integer, Integer> pos = new HashMap<>();
        for (int i = 0; i < n; ++i) pos.put(A[i], i);
        for (int j = 2; j < n; ++j) {
            for (int i = j-1; i > 0; --i) {
                int prev = A[j] - A[i];
                if (prev >= A[i]) break;
                if (!pos.containsKey(prev)) continue;
                dp[i][j] = dp[pos.get(prev)][i] + 1;
                res = Math.max(res, dp[i][j]);
            }
        }
        return res;
    }
}

Analysis:

http://zxi.mytechroad.com/blog/dynamic-programming/leetcode-873-length-of-longest-fibonacci-subsequence/

原文地址：https://www.cnblogs.com/ruruozhenhao/p/10583405.html