Petar Minchev的解释帮助我理清了事情，但我很难解析所有内容，所以我做了一个带有过度描述性变量名和大量注释的Python实现。我做了一个简单的递归解，O(n²)解，和O(n log n)解。

我希望它能帮助理清算法!

递归解决方案

def recursive_solution(remaining_sequence, bigger_than=None):
    """Finds the longest increasing subsequence of remaining_sequence that is      
    bigger than bigger_than and returns it.  This solution is O(2^n)."""

    # Base case: nothing is remaining.                                             
    if len(remaining_sequence) == 0:
        return remaining_sequence

    # Recursive case 1: exclude the current element and process the remaining.     
    best_sequence = recursive_solution(remaining_sequence[1:], bigger_than)

    # Recursive case 2: include the current element if it's big enough.            
    first = remaining_sequence[0]

    if (first > bigger_than) or (bigger_than is None):

        sequence_with = [first] + recursive_solution(remaining_sequence[1:], first)

        # Choose whichever of case 1 and case 2 were longer.                         
        if len(sequence_with) >= len(best_sequence):
            best_sequence = sequence_with

    return best_sequence                                                        

O(n²)动态规划解

def dynamic_programming_solution(sequence):
    """Finds the longest increasing subsequence in sequence using dynamic          
    programming.  This solution is O(n^2)."""

    longest_subsequence_ending_with = []
    backreference_for_subsequence_ending_with = []
    current_best_end = 0

    for curr_elem in range(len(sequence)):
        # It's always possible to have a subsequence of length 1.                    
        longest_subsequence_ending_with.append(1)

        # If a subsequence is length 1, it doesn't have a backreference.             
        backreference_for_subsequence_ending_with.append(None)

        for prev_elem in range(curr_elem):
            subsequence_length_through_prev = (longest_subsequence_ending_with[prev_elem] + 1)

            # If the prev_elem is smaller than the current elem (so it's increasing)   
            # And if the longest subsequence from prev_elem would yield a better       
            # subsequence for curr_elem.                                               
            if ((sequence[prev_elem] < sequence[curr_elem]) and
                    (subsequence_length_through_prev >
                         longest_subsequence_ending_with[curr_elem])):

                # Set the candidate best subsequence at curr_elem to go through prev.    
                longest_subsequence_ending_with[curr_elem] = (subsequence_length_through_prev)
                backreference_for_subsequence_ending_with[curr_elem] = prev_elem
                # If the new end is the best, update the best.    

        if (longest_subsequence_ending_with[curr_elem] >
                longest_subsequence_ending_with[current_best_end]):
            current_best_end = curr_elem
            # Output the overall best by following the backreferences.  

    best_subsequence = []
    current_backreference = current_best_end

    while current_backreference is not None:
        best_subsequence.append(sequence[current_backreference])
        current_backreference = (backreference_for_subsequence_ending_with[current_backreference])

    best_subsequence.reverse()

    return best_subsequence                                                   

O(n log n)动态规划解

def find_smallest_elem_as_big_as(sequence, subsequence, elem):
    """Returns the index of the smallest element in subsequence as big as          
    sequence[elem].  sequence[elem] must not be larger than every element in       
    subsequence.  The elements in subsequence are indices in sequence.  Uses       
    binary search."""

    low = 0
    high = len(subsequence) - 1

    while high > low:
        mid = (high + low) / 2
        # If the current element is not as big as elem, throw out the low half of    
        # sequence.                                                                  
        if sequence[subsequence[mid]] < sequence[elem]:
            low = mid + 1
            # If the current element is as big as elem, throw out everything bigger, but 
        # keep the current element.                                                  
        else:
            high = mid

    return high


def optimized_dynamic_programming_solution(sequence):
    """Finds the longest increasing subsequence in sequence using dynamic          
    programming and binary search (per                                             
    http://en.wikipedia.org/wiki/Longest_increasing_subsequence).  This solution   
    is O(n log n)."""

    # Both of these lists hold the indices of elements in sequence and not the        
    # elements themselves.                                                         
    # This list will always be sorted.                                             
    smallest_end_to_subsequence_of_length = []

    # This array goes along with sequence (not                                     
    # smallest_end_to_subsequence_of_length).  Following the corresponding element 
    # in this array repeatedly will generate the desired subsequence.              
    parent = [None for _ in sequence]

    for elem in range(len(sequence)):
        # We're iterating through sequence in order, so if elem is bigger than the   
        # end of longest current subsequence, we have a new longest increasing          
        # subsequence.                                                               
        if (len(smallest_end_to_subsequence_of_length) == 0 or
                    sequence[elem] > sequence[smallest_end_to_subsequence_of_length[-1]]):
            # If we are adding the first element, it has no parent.  Otherwise, we        
            # need to update the parent to be the previous biggest element.            
            if len(smallest_end_to_subsequence_of_length) > 0:
                parent[elem] = smallest_end_to_subsequence_of_length[-1]
            smallest_end_to_subsequence_of_length.append(elem)
        else:
            # If we can't make a longer subsequence, we might be able to make a        
            # subsequence of equal size to one of our earlier subsequences with a         
            # smaller ending number (which makes it easier to find a later number that 
            # is increasing).                                                          
            # Thus, we look for the smallest element in                                
            # smallest_end_to_subsequence_of_length that is at least as big as elem       
            # and replace it with elem.                                                
            # This preserves correctness because if there is a subsequence of length n 
            # that ends with a number smaller than elem, we could add elem on to the   
            # end of that subsequence to get a subsequence of length n+1.              
            location_to_replace = find_smallest_elem_as_big_as(sequence, smallest_end_to_subsequence_of_length, elem)
            smallest_end_to_subsequence_of_length[location_to_replace] = elem
            # If we're replacing the first element, we don't need to update its parent 
            # because a subsequence of length 1 has no parent.  Otherwise, its parent  
            # is the subsequence one shorter, which we just added onto.                
            if location_to_replace != 0:
                parent[elem] = (smallest_end_to_subsequence_of_length[location_to_replace - 1])

    # Generate the longest increasing subsequence by backtracking through parent.  
    curr_parent = smallest_end_to_subsequence_of_length[-1]
    longest_increasing_subsequence = []

    while curr_parent is not None:
        longest_increasing_subsequence.append(sequence[curr_parent])
        curr_parent = parent[curr_parent]

    longest_increasing_subsequence.reverse()

    return longest_increasing_subsequence         

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)))
  }
}

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

用Java签出包含数组元素的最长递增子序列的代码

http://ideone.com/Nd2eba

/**
 **    Java Program to implement Longest Increasing Subsequence Algorithm
 **/

import java.util.Scanner;

/** Class  LongestIncreasingSubsequence **/
 class  LongestIncreasingSubsequence
{
    /** function lis **/
    public int[] lis(int[] X)
    {        
        int n = X.length - 1;
        int[] M = new int[n + 1];  
        int[] P = new int[n + 1]; 
        int L = 0;

        for (int i = 1; i < n + 1; i++)
        {
            int j = 0;

            /** Linear search applied here. Binary Search can be applied too.
                binary search for the largest positive j <= L such that 
                X[M[j]] < X[i] (or set j = 0 if no such value exists) **/

            for (int pos = L ; pos >= 1; pos--)
            {
                if (X[M[pos]] < X[i])
                {
                    j = pos;
                    break;
                }
            }            
            P[i] = M[j];
            if (j == L || X[i] < X[M[j + 1]])
            {
                M[j + 1] = i;
                L = Math.max(L,j + 1);
            }
        }

        /** backtrack **/

        int[] result = new int[L];
        int pos = M[L];
        for (int i = L - 1; i >= 0; i--)
        {
            result[i] = X[pos];
            pos = P[pos];
        }
        return result;             
    }

    /** Main Function **/
    public static void main(String[] args) 
    {    
        Scanner scan = new Scanner(System.in);
        System.out.println("Longest Increasing Subsequence Algorithm Test\n");

        System.out.println("Enter number of elements");
        int n = scan.nextInt();
        int[] arr = new int[n + 1];
        System.out.println("\nEnter "+ n +" elements");
        for (int i = 1; i <= n; i++)
            arr[i] = scan.nextInt();

        LongestIncreasingSubsequence obj = new LongestIncreasingSubsequence(); 
        int[] result = obj.lis(arr);       

        /** print result **/ 

        System.out.print("\nLongest Increasing Subsequence : ");
        for (int i = 0; i < result.length; i++)
            System.out.print(result[i] +" ");
        System.out.println();
    }
}

好的，我先描述最简单的解也就是O(N²)N是集合的大小。还有一个O(N log N)解，我也会讲到。在高效算法一节中可以找到。

我假设数组的下标从0到N - 1。因此，让我们定义DP[i]为LIS(最长递增子序列)的长度，它结束于索引为i的元素。为了计算DP[i]，我们查看所有索引j < i，并检查DP[j] + 1 > DP[i]和array[j] < array[i](我们希望它是递增的)。如果这是真的，我们可以更新DP[i]的当前最优值。要找到数组的全局最优值，您可以从DP[0…]N - 1]。

int maxLength = 1, bestEnd = 0;
DP[0] = 1;
prev[0] = -1;

for (int i = 1; i < N; i++)
{
   DP[i] = 1;
   prev[i] = -1;

   for (int j = i - 1; j >= 0; j--)
      if (DP[j] + 1 > DP[i] && array[j] < array[i])
      {
         DP[i] = DP[j] + 1;
         prev[i] = j;
      }

   if (DP[i] > maxLength)
   {
      bestEnd = i;
      maxLength = DP[i];
   }
}

我使用数组prev是为了以后能够找到实际的序列，而不仅仅是它的长度。只需在循环中使用prev[bestEnd]从bestEnd递归返回。-1值是停止的标志。

好了，现在来看更有效的O(nlog N)解:

设S[pos]定义为长度为pos的递增序列结束的最小整数。现在遍历输入集的每个整数X，并执行以下操作:

如果X >是S中的最后一个元素，那么将X附加到S的末尾，这本质上意味着我们已经找到了一个新的最大的LIS。 否则，找到S中最小的元素，即>= X，并将其改为X。 因为S在任何时候都是排序的，所以可以使用log(N)的二分搜索来找到元素。

总运行时间- N个整数，并对每个整数进行二进制搜索- N * log(N) = O(N log N)

现在我们来做一个真实的例子:

整数的集合: 2 6 3 4 1 2 9 5 8

步骤:

0. S = {} - Initialize S to the empty set
1. S = {2} - New largest LIS
2. S = {2, 6} - New largest LIS
3. S = {2, 3} - Changed 6 to 3
4. S = {2, 3, 4} - New largest LIS
5. S = {1, 3, 4} - Changed 2 to 1
6. S = {1, 2, 4} - Changed 3 to 2
7. S = {1, 2, 4, 9} - New largest LIS
8. S = {1, 2, 4, 5} - Changed 9 to 5
9. S = {1, 2, 4, 5, 8} - New largest LIS

所以LIS的长度是5 (S的大小)。

为了重建实际的LIS，我们将再次使用父数组。 设parent[i]是LIS中索引为i的元素的前身，该元素以索引为i的元素结束。

为了使事情更简单，我们可以在数组S中保留不是实际的整数，而是它们在集合中的下标(位置)。我们不保留{1,2,4,5,8}，而是保留{4,5,3,7,8}。

即输入[4]= 1，输入[5]= 2，输入[3]= 4，输入[7]= 5，输入[8]= 8。

如果我们正确地更新父数组，实际的LIS是:

input[S[lastElementOfS]], 
input[parent[S[lastElementOfS]]],
input[parent[parent[S[lastElementOfS]]]],
........................................

现在重要的是，我们如何更新父数组?有两种选择:

如果X >是S中的最后一个元素，那么parent[indexX] = indexLastElement。这意味着最新元素的父元素是最后一个元素。我们只是在S的末尾加上X。 否则，找到S中最小元素的索引>= than X，并将其更改为X。这里parent[indexX] = S[index - 1]。