Dynamic programming - 最长单调递增子串问题

关注

原创最新推荐文章于 2026-07-31 08:03:43 发布 513 阅读 AI 写同款· GEO 优化 ›

【C】最长递增子串（LIS）求解 LIS（Longest Increasing Subsequence）最长递增子串问题是一道非常基础的DP问题，这里我们主要是为了给各位解释解决DP问题的最重要的方法之一——记忆化搜索，这里仅讲解复杂度为O（N2）的方法解决此问题，并且这里我们只需要求出最长子串长度，对具体是哪条子串也不做讨论，本文最后会附上O（NlogN）的解决问题的代码，但不做过多的解释说明。我们先把定义弄清楚什么是子串？... 阅读详情

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace LPS
{
    class Program
    {
        static void Main(string[] args)
        {
            int[] pre = { 9, 2, 3, 4, 5, 6 };
            int[,] rec = new int[pre.Length, pre.Length];
            Lps lps = new Lps(rec, pre.Length, pre);
            lps.Lps_initial();
            lps.Lps_length();
            rec = lps.Record;
            int [] res = res = lps.generate ();
            for (int i = 0; i < pre.Length; i++)
            {
                for (int j = 0; j < pre.Length; j++)
                {
                    Console.Write("{0} ", rec[i, j]);
                }
                Console.Write("\n");
            }
            for (int i = res[0]; i <= res[1];i++ )
                Console.Write("{0} ",pre[i]);
            Console.ReadKey();
        }
    }
    class Lps
    {
        public Lps(int [,] record,int x,int [] origin)
        {
            Record = record;
            Length = x;
            OriginSeq = origin;
        }
        public int[,] Record
        {
            get;
            set;
        }
        public int Length
        { set; get; }
        public int[] OriginSeq
        { get; set; }
        public void Lps_initial()
        {
            for (int i = 0; i < Length; i++)
            {
                for (int j = 0; j < Length; j++)
                {
                    if (i == j)
                        Record[i, j] = 1;
                    else
                        Record[i, j] = 0;
                }
            }
        }
        public void Lps_length()
        {
            for (int l = 2; l <= Length; l++)
            {
                for (int i = 0; i < Length - l + 1; i++)
                {
                    int j = i + l - 1;
                    if (Record[i, j - 1] == 0)
                        Record[i, j] = 0;
                    else if (OriginSeq[j - 1] < OriginSeq[j])
                        Record[i, j] = Record[i, j - 1] + 1;
                    else
                        Record[i, j] = 0;
                }
            }
        }
        public int[] generate()
        {
            int temp=0;
            int [] rec = {0,0};
            for (int i = 0; i < Length; i++)
            {
                for (int j = 0; j < Length; j++)
                {
                    if (Record[i, j] > temp)
                    {
                        temp = Record[i, j];
                        rec[0] = i;
                        rec[1] = j;
                    }
                }
            }
            return rec;
        }
    }
}

收录于