LeetCode 873. Length of Longest Fibonacci Subsequence

LeetCode 873. Length of Longest Fibonacci Subsequence


题目描述:

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++.)

题目理解:

给定一个严格增序的整数序列,找到其中满足斐波那切序列的最长子串,最后输出长度。
刚开始的思路就是两个值映射一个值,自然就联想到了2维数组,但是脑袋瓜不灵活,直接用数值本身,导致空间溢出。后面想到变换却忘了影射,用循环,就时间溢出。最后用一个哈希,转化数值本身到坐标,这才解决问题。

class Solution {
public:
    int lenLongestFibSubseq(vector<int>& A) {
        int len = A.size();
        unordered_map<int,int> dict;
        for(int i=0;i<len;i++){
            dict[A[i]] = i;
        }
        vector<vector<int>> store(len,vector<int>(len,2));
        int ans = 0;
        for(int i=0;i<len;i++){
            for(int j=i+1;j<len;j++){
                int sum = A[i] + A[j];
                if(dict.find(sum)!=dict.end()){
                    store[j][dict[sum]] = store[i][j] + 1;
                    if(store[j][dict[sum]] > ans){
                        ans = store[j][dict[sum]];
                    }
                }
            }    
        }
        return ans;

    }
};

猜你喜欢

转载自blog.csdn.net/qq_40280096/article/details/81838240
今日推荐