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
说到DP solution，我发现很奇怪的是没有人提到LIS可以简化为LCS。你所需要做的就是对原始序列的副本进行排序，删除所有重复的副本，然后对它们进行LCS。在伪代码中是:
def LIS(S):
T = sort(S)
T = removeDuplicates(T)
return LCS(S, T)
以及用Go语言编写的完整实现。如果你不需要重构解，你就不需要维护整个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]
}
其他回答
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
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。
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i将表示的元素，A =以i结尾的最长递增子序列的长度，B =以列表[i]结尾的最长递增子序列中列表[i]的前身的索引。
从元素1开始，元素1的元组列表为[(1,0)] 对于元素i，扫描列表0..i，找到元素list[k]，使得list[k] < list[i]，元素i的A值，Ai为Ak + 1, Bi为k。如果有多个这样的元素，将它们添加到元素i的元组列表中。
最后，找到所有最大值为A (LIS以element结尾的长度)的元素，并使用元组回溯以获得列表。
我已经在http://www.edufyme.com/code/?id=66f041e16a60928b05a7e228a89c3799上分享了相同的代码
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。
说到DP solution，我发现很奇怪的是没有人提到LIS可以简化为LCS。你所需要做的就是对原始序列的副本进行排序，删除所有重复的副本，然后对它们进行LCS。在伪代码中是:
def LIS(S):
T = sort(S)
T = removeDuplicates(T)
return LCS(S, T)
以及用Go语言编写的完整实现。如果你不需要重构解，你就不需要维护整个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]
}
