我有一组整数。我想使用动态编程找到该集合的longest increasing subsequence。
【问题讨论】：
标签： algorithm computer-science dynamic-programming memoization lis
好的，我将首先描述最简单的解决方案，即 O(N^2)，其中 N 是集合的大小。还有一个 O(N log N) 的解决方案，我也会描述。在高效算法部分查看here。
我将假设数组的索引是从 0 到 N - 1。所以让我们将 DP[i] 定义为 LIS（最长递增子序列）的长度，它在索引为 i 的元素处结束。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]（我们希望它增加）。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] 中获取最大值。
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 以便以后能够找到实际序列，而不仅仅是它的长度。只需使用prev[bestEnd] 在循环中从bestEnd 递归返回即可。 -1 值是停止的标志。
O(N log N) 解决方案：让S[pos] 定义为结束长度递增序列pos 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 > S 中的最后一个元素，则将X 附加到S 的末尾。这实质上意味着我们找到了一个新的最大的LIS。
否则找到S中最小的元素，即>=比X，改成X。
因为S是随时排序的，所以可以在log(N)中使用二分查找找到元素。
总运行时间 - N 整数和对它们中的每一个的二进制搜索 - N * log(N) = O(N log N)
现在让我们做一个真实的例子：
整数集合：
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）。
为了重构实际的LIS，我们将再次使用父数组。
让parent[i] 成为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。
为了简单起见，我们可以在数组S 中保存，而不是实际的整数，而是它们在集合中的索引（位置）。我们不保留{1, 2, 4, 5, 8}，而是保留{4, 5, 3, 7, 8}。
即input[4] = 1, input[5] = 2, input[3] = 4, input[7 ] = 5，输入[8] = 8。
如果我们正确更新父数组，实际的 LIS 是：
input[S[lastElementOfS]],
input[parent[S[lastElementOfS]]],
input[parent[parent[S[lastElementOfS]]]],
........................................
现在重要的事情——我们如何更新父数组？有两种选择：
如果X > 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 的末尾添加X。
否则找到S中最小元素的索引，即>=而不是X，改成X。这里parent[indexX] = S[index - 1]。
【讨论】：
DP[j] + 1 == DP[i] 那么DP[i] 不会因为DP[i] = DP[j] + 1 而变得更好。我们正在尝试优化DP[i]。
[1,2,5,8]，数组中4在1之前，LIS怎么可能是[1,2,4,5,8]？
[2,3,4,5,8]。仔细阅读 - S 数组 DOES NOT 代表一个实际序列。 Let S[pos] be defined as the smallest integer that ends an increasing sequence of length pos.
Petar Minchev 的解释帮助我理清了问题，但我很难解析所有内容，因此我使用过度描述的变量名称和大量 cmets 进行了 Python 实现。我做了一个朴素的递归解决方案，O(n^2) 解决方案和 O(n log n) 解决方案。
我希望它有助于清理算法！
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
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
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
【讨论】：
bisect 可能是有意义的。为了演示算法的工作原理及其性能特征，我尽量保持原始。
说到DP解决方案，我很惊讶没有人提到LIS可以简化为LCS。您需要做的就是对原始序列的副本进行排序，删除所有重复项并对它们进行 LCS。在伪代码中是：
def LIS(S):
T = sort(S)
T = removeDuplicates(T)
return LCS(S, T)
以及用 Go 编写的完整实现。如果不需要重构解，则不需要维护整个 n^2 DP 矩阵。
func lcs(arr1 []int) int {
arr2 := make([]int, len(arr1))
for i, v := range arr1 {
arr2[i] = v
}
sort.Ints(arr1)
arr3 := []int{}
prev := arr1[0] - 1
for _, v := range arr1 {
if v != prev {
prev = v
arr3 = append(arr3, v)
}
}
n1, n2 := len(arr1), len(arr3)
M := make([][]int, n2 + 1)
e := make([]int, (n1 + 1) * (n2 + 1))
for i := range M {
M[i] = e[i * (n1 + 1):(i + 1) * (n1 + 1)]
}
for i := 1; i <= n2; i++ {
for j := 1; j <= n1; j++ {
if arr2[j - 1] == arr3[i - 1] {
M[i][j] = M[i - 1][j - 1] + 1
} else if M[i - 1][j] > M[i][j - 1] {
M[i][j] = M[i - 1][j]
} else {
M[i][j] = M[i][j - 1]
}
}
}
return M[n2][n1]
}
【讨论】：
以下 C++ 实现还包括一些代码，这些代码使用名为 prev 的数组构建实际的最长递增子序列。
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;
}
【讨论】：
以下是从动态规划的角度评估问题的三个步骤：
如果我们以序列 {0, 8, 2, 3, 7, 9} 为例，在索引处：
这是 C++11 的工作代码：
#include <iostream>
#include <vector>
int getLongestIncSub(const std::vector<int> &sequence, size_t index, std::vector<std::vector<int>> &sub) {
if(index == 0) {
sub.push_back(std::vector<int>{sequence[0]});
return 1;
}
size_t longestSubSeq = getLongestIncSub(sequence, index - 1, sub);
std::vector<std::vector<int>> tmpSubSeq;
for(std::vector<int> &subSeq : sub) {
if(subSeq[subSeq.size() - 1] < sequence[index]) {
std::vector<int> newSeq(subSeq);
newSeq.push_back(sequence[index]);
longestSubSeq = std::max(longestSubSeq, newSeq.size());
tmpSubSeq.push_back(newSeq);
}
}
std::copy(tmpSubSeq.begin(), tmpSubSeq.end(),
std::back_insert_iterator<std::vector<std::vector<int>>>(sub));
return longestSubSeq;
}
int getLongestIncSub(const std::vector<int> &sequence) {
std::vector<std::vector<int>> sub;
return getLongestIncSub(sequence, sequence.size() - 1, sub);
}
int main()
{
std::vector<int> seq{0, 8, 2, 3, 7, 9};
std::cout << getLongestIncSub(seq);
return 0;
}
【讨论】：
这是 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)))
}
}
【讨论】：
这是另一个 O(n^2) 的 JAVA 实现。没有递归/记忆来生成实际的子序列。只是一个存储每个阶段的实际 LIS 的字符串数组和一个存储每个元素的 LIS 长度的数组。很简单。看看：
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);
}
}
【讨论】：
这里是 java O(nlogn) 实现
import java.util.Scanner;
public class LongestIncreasingSeq {
private static int binarySearch(int table[],int a,int len){
int end = len-1;
int beg = 0;
int mid = 0;
int result = -1;
while(beg <= end){
mid = (end + beg) / 2;
if(table[mid] < a){
beg=mid+1;
result = mid;
}else if(table[mid] == a){
return len-1;
}else{
end = mid-1;
}
}
return result;
}
public static void main(String[] args) {
// int[] t = {1, 2, 5,9,16};
// System.out.println(binarySearch(t , 9, 5));
Scanner in = new Scanner(System.in);
int size = in.nextInt();//4;
int A[] = new int[size];
int table[] = new int[A.length];
int k = 0;
while(k<size){
A[k++] = in.nextInt();
if(k<size-1)
in.nextLine();
}
table[0] = A[0];
int len = 1;
for (int i = 1; i < A.length; i++) {
if(table[0] > A[i]){
table[0] = A[i];
}else if(table[len-1]<A[i]){
table[len++]=A[i];
}else{
table[binarySearch(table, A[i],len)+1] = A[i];
}
}
System.out.println(len);
}
}
//可以使用TreeSet
【讨论】：
这可以使用动态编程在 O(n^2) 内解决。相同的 Python 代码如下：-
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 处的值表示该 list_index 的最长递增子序列长度。
【讨论】：
这是一个 O(n^2) 的 Java 实现。我只是没有使用二进制搜索来查找 S 中的最小元素，即 >= 比 X。我只是使用了一个 for 循环。使用二分搜索会使复杂度达到 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));
}
【讨论】：
检查java中的代码以获得数组元素的最长递增子序列
/**
** 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();
}
}
【讨论】：
这可以使用动态规划在 O(n^2) 内解决。
按顺序处理输入元素并为每个元素维护一个元组列表。每个元组 (A,B)，对于元素 i 将表示，A = 以 i 结尾的最长递增子序列的长度，B = list[i] 在以 list[i 结尾的最长递增子序列中的前任索引].
从元素 1 开始，元素 1 的元组列表将是 [(1,0)] 对于元素 i，扫描列表 0..i 并找到元素 list[k] 使得 list[k]
最后找到最大值为A的所有元素（LIS在元素处结束的长度），并使用元组回溯得到列表。
我已经在http://www.edufyme.com/code/?id=66f041e16a60928b05a7e228a89c3799分享了相同的代码
【讨论】：
O(n^2) 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]+" ");
}
}
【讨论】：
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（这在 O(n^2) 时间内解决），但这种方式仍然提供了很好的动态编程方法。
【讨论】：
这是我使用二分搜索的 Leetcode 解决方案：->
class Solution:
def binary_search(self,s,x):
low=0
high=len(s)-1
flag=1
while low<=high:
mid=(high+low)//2
if s[mid]==x:
flag=0
break
elif s[mid]<x:
low=mid+1
else:
high=mid-1
if flag:
s[low]=x
return s
def lengthOfLIS(self, nums: List[int]) -> int:
if not nums:
return 0
s=[]
s.append(nums[0])
for i in range(1,len(nums)):
if s[-1]<nums[i]:
s.append(nums[i])
else:
s=self.binary_search(s,nums[i])
return len(s)
【讨论】：
时间复杂度为 O(nlog(n)) 的 C++ 中最简单的 LIS 解决方案
#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
【讨论】：
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+" ");
}
}
}
【讨论】：
我已经使用动态编程和记忆化在 java 中实现了 LIS。除了代码之外，我还进行了复杂度计算，即为什么它是 O(n Log(base2) n)。因为我觉得理论上或逻辑上的解释很好，但实际演示总是更容易理解。
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时 -
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
【讨论】：
寻找最长递增子序列的 O(NLog(N)) 方法
让我们维护一个数组，其中第 i 个元素是 i 大小的子序列可以结束的最小可能数。
我故意避免更多细节，因为投票最多的答案已经解释了它，但这种技术最终会导致使用集合数据结构（至少在 c++ 中）的简洁实现。
这是c++中的实现（假设需要严格增加最长子序列大小）
#include <bits/stdc++.h> // gcc supported header to include (almost) everything
using namespace std;
typedef long long ll;
int main()
{
ll n;
cin >> n;
ll arr[n];
set<ll> S;
for(ll i=0; i<n; i++)
{
cin >> arr[i];
auto it = S.lower_bound(arr[i]);
if(it != S.end())
S.erase(it);
S.insert(arr[i]);
}
cout << S.size() << endl; // Size of the set is the required answer
return 0;
}
【讨论】：
说明
此算法涉及创建节点格式为(a,b) 的树。
a 表示目前我们正在考虑附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 被附加到我们目前拥有的子数组的末尾，则将根据该索引做出下一个决定。
算法
我们从一个无效的根 (INT_MIN,0) 开始，指向数组的索引零，因为此时子序列为空，即b = 0。
Base Case：如果b >= array.length，则返回1。
循环遍历数组中从b 索引到数组末尾的所有元素，即i = b ... array.length-1。
i) 如果元素array[i] 是greater than 当前a，则它有资格被认为是要附加到我们目前拥有的子序列的元素之一。
ii) 递归到节点(array[i],b+1)，其中a 是我们在2(i) 中遇到的元素，可以附加到我们目前的子序列中。 b+1 是要考虑的数组的下一个索引。
iii) 返回通过i = b ... array.length 循环获得的max 长度。如果a 大于i = b to array.length 中的任何其他元素，则返回1。
计算构建为level 的树的级别。最后，level - 1 是所需的LIS。那是树的最长路径中edges 的数量。
注意：算法的记忆部分被省略了，因为它从树中清晰可见。
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;
}
【讨论】：