我有一组整数。我想用动态规划的方法求出这个集合的最长递增子序列。
当前回答
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
其他回答
c++中最简单的LIS解决方案，具有O(nlog(n))时间复杂度
#include <iostream>
#include "vector"
using namespace std;
// binary search (If value not found then it will return the index where the value should be inserted)
int ceilBinarySearch(vector<int> &a,int beg,int end,int value)
{
if(beg<=end)
{
int mid = (beg+end)/2;
if(a[mid] == value)
return mid;
else if(value < a[mid])
return ceilBinarySearch(a,beg,mid-1,value);
else
return ceilBinarySearch(a,mid+1,end,value);
return 0;
}
return beg;
}
int lis(vector<int> arr)
{
vector<int> dp(arr.size(),0);
int len = 0;
for(int i = 0;i<arr.size();i++)
{
int j = ceilBinarySearch(dp,0,len-1,arr[i]);
dp[j] = arr[i];
if(j == len)
len++;
}
return len;
}
int main()
{
vector<int> arr {2, 5,-1,0,6,1,2};
cout<<lis(arr);
return 0;
}
输出: 4
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
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]+" ");
}
}
最长递增子序列(Java)
import java.util.*;
class ChainHighestValue implements Comparable<ChainHighestValue>{
int highestValue;
int chainLength;
ChainHighestValue(int highestValue,int chainLength) {
this.highestValue = highestValue;
this.chainLength = chainLength;
}
@Override
public int compareTo(ChainHighestValue o) {
return this.chainLength-o.chainLength;
}
}
public class LongestIncreasingSubsequenceLinkedList {
private static LinkedList<Integer> LongestSubsequent(int arr[], int size){
ArrayList<LinkedList<Integer>> seqList=new ArrayList<>();
ArrayList<ChainHighestValue> valuePairs=new ArrayList<>();
for(int i=0;i<size;i++){
int currValue=arr[i];
if(valuePairs.size()==0){
LinkedList<Integer> aList=new LinkedList<>();
aList.add(arr[i]);
seqList.add(aList);
valuePairs.add(new ChainHighestValue(arr[i],1));
}else{
try{
ChainHighestValue heighestIndex=valuePairs.stream().filter(e->e.highestValue<currValue).max(ChainHighestValue::compareTo).get();
int index=valuePairs.indexOf(heighestIndex);
seqList.get(index).add(arr[i]);
heighestIndex.highestValue=arr[i];
heighestIndex.chainLength+=1;
}catch (Exception e){
LinkedList<Integer> aList=new LinkedList<>();
aList.add(arr[i]);
seqList.add(aList);
valuePairs.add(new ChainHighestValue(arr[i],1));
}
}
}
ChainHighestValue heighestIndex=valuePairs.stream().max(ChainHighestValue::compareTo).get();
int index=valuePairs.indexOf(heighestIndex);
return seqList.get(index);
}
public static void main(String[] args){
int arry[]={5,1,3,6,11,30,32,5,3,73,79};
//int arryB[]={3,1,5,2,6,4,9};
LinkedList<Integer> LIS=LongestSubsequent(arry, arry.length);
System.out.println("Longest Incrementing Subsequence:");
for(Integer a: LIS){
System.out.print(a+" ");
}
}
}
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。
