[leetcode]873. Length of Longest Fibonacci Subsequence

Analysis

中午吃啥—— [每天刷題並不難0.0]

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].)

Explanation:

第一種方法時間複雜度是O(n^2logm)， m=A[i]+A[j]。
第二種方法動歸解決，時間複雜度是O(n^2)。

Implement

方法一：

class Solution {
public:
    int lenLongestFibSubseq(vector<int>& A) {
        unordered_set<int> nums(A.begin(), A.end());
        int res = 0;
        int len;
        int a, b;
        for(int i=0; i<A.size(); i++){
            for(int j=i+1; j<A.size(); j+
 
+){
                a = A[i];
                b = A[j];
                len = 2;
                while(nums.count(a+b)){
                    b = a+b;
                    a = b-a;
                    len++;
                }
                res = max(res, len);
            }
        }
        return res>2?res:0;
    }
};

方法二：

class Solution {
public:
    int lenLongestFibSubseq(vector<int>& A) {
        int len = A.size();
        int res = 0;
        vector<vector<int>> dp(len, vector<int>(len, 2));
        map<int, int> idx;
        for(int i=0; i<len; i++)
            idx[A[i]] = i;
        for(int j=1; j<len; j++){
            for(int i=0; i<j; i++){
                if(idx.find(A[j]-A[i]) != idx.end() && idx[A[j]-A[i]] < i){
                    int tmp = idx[A[j]-A[i]];
                    dp[i][j] = max(dp[i][j], dp[tmp][i]+1);
                }
            }
        }
        for(int i=0; i<len; i++){
            for(int j=i+1; j<len; j++)
                res = max(res, dp[i][j]);
        }
        return res>2?res:0;
    }
};