下面的c++实现还包括一些使用名为prev的数组构建实际最长递增子序列的代码。

std::vector<int> longest_increasing_subsequence (const std::vector<int>& s)
{
    int best_end = 0;
    int sz = s.size();

    if (!sz)
        return std::vector<int>();

    std::vector<int> prev(sz,-1);
    std::vector<int> memo(sz, 0);

    int max_length = std::numeric_limits<int>::min();

    memo[0] = 1;

    for ( auto i = 1; i < sz; ++i)
    {
        for ( auto j = 0; j < i; ++j)
        {
            if ( s[j] < s[i] && memo[i] < memo[j] + 1 )
            {
                memo[i] =  memo[j] + 1;
                prev[i] =  j;
            }
        }

        if ( memo[i] > max_length ) 
        {
            best_end = i;
            max_length = memo[i];
        }
    }

    // Code that builds the longest increasing subsequence using "prev"
    std::vector<int> results;
    results.reserve(sz);

    std::stack<int> stk;
    int current = best_end;

    while (current != -1)
    {
        stk.push(s[current]);
        current = prev[current];
    }

    while (!stk.empty())
    {
        results.push_back(stk.top());
        stk.pop();
    }

    return results;
}

没有堆栈的实现只是反转向量

#include <iostream>
#include <vector>
#include <limits>
std::vector<int> LIS( const std::vector<int> &v ) {
  auto sz = v.size();
  if(!sz)
    return v;
  std::vector<int> memo(sz, 0);
  std::vector<int> prev(sz, -1);
  memo[0] = 1;
  int best_end = 0;
  int max_length = std::numeric_limits<int>::min();
  for (auto i = 1; i < sz; ++i) {
    for ( auto j = 0; j < i ; ++j) {
      if (s[j] < s[i] && memo[i] < memo[j] + 1) {
        memo[i] = memo[j] + 1;
        prev[i] = j;
      }
    }
    if(memo[i] > max_length) {
      best_end = i;
      max_length = memo[i];
    }
  }

  // create results
  std::vector<int> results;
  results.reserve(v.size());
  auto current = best_end;
  while (current != -1) {
    results.push_back(s[current]);
    current = prev[current];
  }
  std::reverse(results.begin(), results.end());
  return results;
}

这是另一个O(n²)JAVA实现。不需要递归/记忆来生成实际的子序列。只是一个字符串数组，存储每个阶段的实际LIS和一个数组，存储每个元素的LIS的长度。非常简单。看看吧:

import java.io.BufferedReader;
import java.io.InputStreamReader;

/**
 * Created by Shreyans on 4/16/2015
 */

class LNG_INC_SUB//Longest Increasing Subsequence
{
    public static void main(String[] args) throws Exception
    {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        System.out.println("Enter Numbers Separated by Spaces to find their LIS\n");
        String[] s1=br.readLine().split(" ");
        int n=s1.length;
        int[] a=new int[n];//Array actual of Numbers
        String []ls=new String[n];// Array of Strings to maintain LIS for every element
        for(int i=0;i<n;i++)
        {
            a[i]=Integer.parseInt(s1[i]);
        }
        int[]dp=new int[n];//Storing length of max subseq.
        int max=dp[0]=1;//Defaults
        String seq=ls[0]=s1[0];//Defaults
        for(int i=1;i<n;i++)
        {
            dp[i]=1;
            String x="";
            for(int j=i-1;j>=0;j--)
            {
                //First check if number at index j is less than num at i.
                // Second the length of that DP should be greater than dp[i]
                // -1 since dp of previous could also be one. So we compare the dp[i] as empty initially
                if(a[j]<a[i]&&dp[j]>dp[i]-1)
                {
                    dp[i]=dp[j]+1;//Assigning temp length of LIS. There may come along a bigger LIS of a future a[j]
                    x=ls[j];//Assigning temp LIS of a[j]. Will append a[i] later on
                }
            }
            x+=(" "+a[i]);
            ls[i]=x;
            if(dp[i]>max)
            {
                max=dp[i];
                seq=ls[i];
            }
        }
        System.out.println("Length of LIS is: " + max + "\nThe Sequence is: " + seq);
    }
}

实际代码:http://ideone.com/sBiOQx

这是一个O(n²)的Java实现。我只是没有使用二分搜索来找到S中最小的元素，它是>= than x，我只是使用了一个for循环。使用二分搜索将使复杂度为O(n logn)

public static void olis(int[] seq){

    int[] memo = new int[seq.length];

    memo[0] = seq[0];
    int pos = 0;

    for (int i=1; i<seq.length; i++){

        int x = seq[i];

            if (memo[pos] < x){ 
                pos++;
                memo[pos] = x;
            } else {

                for(int j=0; j<=pos; j++){
                    if (memo[j] >= x){
                        memo[j] = x;
                        break;
                    }
                }
            }
            //just to print every step
            System.out.println(Arrays.toString(memo));
    }

    //the final array with the LIS
    System.out.println(Arrays.toString(memo));
    System.out.println("The length of lis is " + (pos + 1));

}

我已经在java中使用动态编程和记忆实现了LIS。随着代码，我做了复杂性计算，即为什么它是O(n Log(base2) n)。因为我觉得理论或逻辑解释是很好的，但实际演示总是更好的理解。

package com.company.dynamicProgramming;

import java.util.HashMap;
import java.util.Map;

public class LongestIncreasingSequence {

    static int complexity = 0;

    public static void main(String ...args){


        int[] arr = {10, 22, 9, 33, 21, 50, 41, 60, 80};
        int n = arr.length;

        Map<Integer, Integer> memo = new HashMap<>();

        lis(arr, n, memo);

        //Display Code Begins
        int x = 0;
        System.out.format("Longest Increasing Sub-Sequence with size %S is -> ",memo.get(n));
        for(Map.Entry e : memo.entrySet()){

            if((Integer)e.getValue() > x){
                System.out.print(arr[(Integer)e.getKey()-1] + " ");
                x++;
            }
        }
        System.out.format("%nAnd Time Complexity for Array size %S is just %S ", arr.length, complexity );
        System.out.format( "%nWhich is equivalent to O(n Log n) i.e. %SLog(base2)%S is %S",arr.length,arr.length, arr.length * Math.ceil(Math.log(arr.length)/Math.log(2)));
        //Display Code Ends

    }



    static int lis(int[] arr, int n, Map<Integer, Integer> memo){

        if(n==1){
            memo.put(1, 1);
            return 1;
        }

        int lisAti;
        int lisAtn = 1;

        for(int i = 1; i < n; i++){
            complexity++;

            if(memo.get(i)!=null){
                lisAti = memo.get(i);
            }else {
                lisAti = lis(arr, i, memo);
            }

            if(arr[i-1] < arr[n-1] && lisAti +1 > lisAtn){
                lisAtn = lisAti +1;
            }
        }

        memo.put(n, lisAtn);
        return lisAtn;

    }
}

当我运行上面的代码-

Longest Increasing Sub-Sequence with size 6 is -> 10 22 33 50 60 80 
And Time Complexity for Array size 9 is just 36 
Which is equivalent to O(n Log n) i.e. 9Log(base2)9 is 36.0
Process finished with exit code 0

这可以用动态规划在O(n²)中解决。同样的Python代码是这样的:-

def LIS(numlist):
    LS = [1]
    for i in range(1, len(numlist)):
        LS.append(1)
        for j in range(0, i):
            if numlist[i] > numlist[j] and LS[i]<=LS[j]:
                LS[i] = 1 + LS[j]
    print LS
    return max(LS)

numlist = map(int, raw_input().split(' '))
print LIS(numlist)

输入:5 19 5 81 50 28 29 1 83 23

输出将是:[1,2,1,3,3,3,4,1,5,3] 5

输出列表的list_index是输入列表的list_index。输出列表中给定list_index的值表示该list_index的最长递增子序列长度。

下面是O(n^2)算法的Scala实现:

object Solve {
  def longestIncrSubseq[T](xs: List[T])(implicit ord: Ordering[T]) = {
    xs.foldLeft(List[(Int, List[T])]()) {
      (sofar, x) =>
        if (sofar.isEmpty) List((1, List(x)))
        else {
          val resIfEndsAtCurr = (sofar, xs).zipped map {
            (tp, y) =>
              val len = tp._1
              val seq = tp._2
              if (ord.lteq(y, x)) {
                (len + 1, x :: seq) // reversely recorded to avoid O(n)
              } else {
                (1, List(x))
              }
          }
          sofar :+ resIfEndsAtCurr.maxBy(_._1)
        }
    }.maxBy(_._1)._2.reverse
  }

  def main(args: Array[String]) = {
    println(longestIncrSubseq(List(
      0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15)))
  }
}