Longest monotonous consequent sub-sequence_longest increase sub sequence-优快云博客

本文链接：https://blog.youkuaiyun.com/u012212811/article/details/45333695

本文提供了一种经典的解决方案，通过从给定序列的左侧运行，判断相邻元素的大小关系来找到最长连续子序列。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Problem: Find out the longest continuous sub-sequence.

Analysis:

This is a canonical solution. Runing from the ledt side of the given sequence, if the i+1th element is bigger or equal to the ith one, add to a temporary vector. Otherwise, the i+1th element will not increase the length of the current stored sub-sequence. Then, compare the temporary vector with the returned vector, replace the returned vector with the temporary one.

The time complexity is O(n).

Solutions:

C++:

vector<int> longest_monotonous_sequence(int seq[], int n, bool isIncreased = true)
{
    vector<int> lms;
    if(n == 0)
        return lms;

    vector<int> temp_mon_seq;
    for(int i = 0; i < n; ++i) {
        if(temp_mon_seq.empty())
            temp_mon_seq.push_back(seq[i]);
        else {
            if(isIncreased) {
                if(seq[i] >= *(temp_mon_seq.end() - 1))
                    temp_mon_seq.push_back(seq[i]);
                else {
                    if(temp_mon_seq.size() > lms.size())
                        lms = temp_mon_seq;
                    temp_mon_seq.clear();
                }
            } else {
                if(seq[i] <= *(temp_mon_seq.end() - 1))
                    temp_mon_seq.push_back(seq[i]);
                else {
                    if(temp_mon_seq.size() > lms.size())
                        lms = temp_mon_seq;
                    temp_mon_seq.clear();
                }
            }
        }
    }

    if(temp_mon_seq.size() > lms.size())
        lms = temp_mon_seq;

    return lms;
}