LeetCode-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++.)

题解：

第一次O(n^3)直接tle

class Solution {
public:
  int lenLongestFibSubseq(vector<int>& A) {
    int n = A.size();
    int maxi = 0;
    for (int i = 0; i < n; i++) {
      for (int j = i + 1; j < n; j++) {
        int len = 0;
        queue<int> q;
        q.push(A[i]);
        q.push(A[j]);
        for (int k = j + 1; k < n; k++) {
          if (q.front() + q.back() == A[k]) {
            q.pop();
            q.push(A[k]);
            if (len == 0) {
              len = 3;
            }
            else {
              len++;
            }
          }
        }
        maxi = max(len, maxi);
      }
    }
    return maxi;
  }
};

第二次vector查找仍然tle

class Solution {
public:
  int lenLongestFibSubseq(vector<int>& A) {
    int n = A.size();
    int maxi = 0;
    for (int i = 0; i < n; i++) {
      for (int j = i + 1; j < n; j++) {
        int len = 0;
        int a = A[i];
        int b = A[j];
        vector<int>::iterator tmp = find(A.begin(), A.end(), a + b);
        while(tmp != A.end()) {
          b = a + b;
          a = b - a;
          if (len == 0) {
            len = 3;
          }
          else {
            len++;
          }
          tmp = find(A.begin(), A.end(), a + b);
        }
        maxi = max(maxi, len);
      }
    }
    return maxi;
  }
};

只能用map了啊，AC

class Solution {
public:
  int lenLongestFibSubseq(vector<int>& A) {
    int n = A.size();
    map<int, int> mp;
    int maxi = 0;
    for (int i = 0; i < n; i++) {
      mp.insert(pair<int, int>(A[i], 1));
    }
    for (int i = 0; i < n; i++) {
      for (int j = i + 1; j < n; j++) {
        int len = 0;
        int a = A[i], b = A[j];
        while (true) {
          if (mp.find(a + b) != mp.end()) {
            b = a + b;
            a = b - a;
            if (len == 0) {
              len = 3;
            }
            else {
              len++;
            }
          }
          else {
            break;
          }
        }
        maxi = max(maxi, len);
      }
    }
    return maxi;
  }
};