求最长递增子序列(LIS)的O(NLog(N))递归DP方法

解释

该算法涉及创建节点格式为(a,b)的树。

A表示到目前为止我们考虑添加到有效子序列的下一个元素。

B表示剩余子数组的起始索引，如果a被添加到目前为止我们所拥有的子数组的末尾，则下一个决策将从该子数组开始。

算法

We start with an invalid root (INT_MIN,0), pointing at index zero of the array since subsequence is empty at this point, i.e. b = 0. Base Case: return 1 if b >= array.length. Loop through all the elements in the array from the b index to the end of the array, i.e i = b ... array.length-1. i) If an element, array[i] is greater than the current a, it is qualified to be considered as one of the elements to be appended to the subsequence we have so far. ii) Recurse into the node (array[i],b+1), where a is the element we encountered in 2(i) which is qualified to be appended to the subsequence we have so far. And b+1 is the next index of the array to be considered. iii) Return the max length obtained by looping through i = b ... array.length. In a case where a is bigger than any other element from i = b to array.length, return 1. Compute the level of the tree built as level. Finally, level - 1 is the desired LIS. That is the number of edges in the longest path of the tree.

注意:算法的记忆部分被省略了，因为它是从树中清除的。

随便举个例子 标记为x的节点从DB内存值中获取。

Java实现

public int lengthOfLIS(int[] nums) {
            return LIS(nums,Integer.MIN_VALUE, 0,new HashMap<>()) -1;
    }
    public int LIS(int[] arr, int value, int nextIndex, Map<String,Integer> memo){
        if(memo.containsKey(value+","+nextIndex))return memo.get(value+","+nextIndex);
        if(nextIndex >= arr.length)return 1;

        int max = Integer.MIN_VALUE;
        for(int i=nextIndex; i<arr.length; i++){
            if(arr[i] > value){
                max = Math.max(max,LIS(arr,arr[i],i+1,memo));
            }
        }
        if(max == Integer.MIN_VALUE)return 1;
        max++;
        memo.put(value+","+nextIndex,max);
        return max;
    }

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

def longestincrsub(arr1):
    n=len(arr1)
    l=[1]*n
    for i in range(0,n):
        for j in range(0,i)  :
            if arr1[j]<arr1[i] and l[i]<l[j] + 1:
                l[i] =l[j] + 1
    l.sort()
    return l[-1]
arr1=[10,22,9,33,21,50,41,60]
a=longestincrsub(arr1)
print(a)

尽管有一种方法可以在O(nlogn)时间内解决这个问题(它在O(n²)时间内解决)但这种方法仍然提供了动态规划方法，这也是很好的。

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)))
  }
}

O(n²)java实现:

void LIS(int arr[]){
        int maxCount[]=new int[arr.length];
        int link[]=new int[arr.length];
        int maxI=0;
        link[0]=0;
        maxCount[0]=0;

        for (int i = 1; i < arr.length; i++) {
            for (int j = 0; j < i; j++) {
                if(arr[j]<arr[i] && ((maxCount[j]+1)>maxCount[i])){
                    maxCount[i]=maxCount[j]+1;
                    link[i]=j;
                    if(maxCount[i]>maxCount[maxI]){
                        maxI=i;
                    }
                }
            }
        }


        for (int i = 0; i < link.length; i++) {
            System.out.println(arr[i]+"   "+link[i]);
        }
        print(arr,maxI,link);

    }

    void print(int arr[],int index,int link[]){
        if(link[index]==index){
            System.out.println(arr[index]+" ");
            return;
        }else{
            print(arr, link[index], link);
            System.out.println(arr[index]+" ");
        }
    }