给定由N个整数和整数K组成的数组arr [] ，任务是找到总和等于K的最长子序列的长度。
例子：
Input: arr[] = {-4, -2, -2, -1, 6}, K = 0
Output: 3
Explanation:
The longest subsequence is of length 3 which is {-4, -2, 6} having sum 0.
Input: arr[] = {-3, 0, 1, 1, 2}, K = 1
Output: 5
Explanation: The longest subsequence is of length 5 which is {-3, 0, 1, 1, 2} having sum 1.
天真的方法：解决问题的最简单方法是生成所有可能的不同长度的子序列，并检查它们的总和是否等于K。在所有这些具有和K的子序列中，找到长度最长的子序列。
时间复杂度： O(2 N )
递归和回溯方法：此问题的基本方法是对向量进行排序，找到所有可能的子序列的和，然后拾取具有给定总和的最大长度的子序列。这可以使用递归和回溯来完成。
请按照以下步骤解决此问题：
- 对给定的数组/向量进行排序。
- 将全局变量max_length初始化为0，该变量存储最大长度子集。
- 对于数组中的每个索引i ，调用递归函数以找出所有可能的子集，这些子集具有[i，N-1]范围内的元素之和为K。
- 每次找到总和为K的子集时，请检查其大小是否大于当前的max_length值。如果是，则更新max_length的值。
- 计算完所有可能的子集和之后，返回max_length 。
下面是上述方法的实现：
C++
// C++ Program to implement the
// above approach
#include
using namespace std;
// Initialise maximum possible
// length of subsequence
int max_length = 0;
// Store elements to compare
// max_length with its size
// and change the value of
// max_length accordingly
vector store;
// Store the elements of the
// longest subsequence
vector ans;
// Function to find the length
// of longest subsequence
void find_max_length(
vector& arr,
int index, int sum, int k)
{
sum = sum + arr[index];
store.push_back(arr[index]);
if (sum == k) {
if (max_length < store.size()) {
// Update max_length
max_length = store.size();
// Store the subsequence
// elements
ans = store;
}
}
for (int i = index + 1;
i < arr.size(); i++) {
if (sum + arr[i] <= k) {
// Recursively proceed
// with obtained sum
find_max_length(arr, i,
sum, k);
// poping elements
// from back
// of vector store
store.pop_back();
}
// if sum > 0 then we don't
// required thatsubsequence
// so return and continue
// with earlier elements
else
return;
}
return;
}
int longestSubsequence(vector arr,
int n, int k)
{
// Sort the given array
sort(arr.begin(), arr.end());
// Traverse the array
for (int i = 0; i < n; i++) {
// If max_length is already
// greater than or equal
// than remaining length
if (max_length >= n - i)
break;
store.clear();
find_max_length(arr, i, 0, k);
}
return max_length;
}
// Driver code
int main()
{
vector arr{ -3, 0, 1, 1, 2 };
int n = arr.size();
int k = 1;
cout << longestSubsequence(arr,
n, k);
return 0;
} Java
// Java Program to implement the
// above approach
import java.util.*;
class GFG{
// Initialise maximum possible
// length of subsequence
static int max_length = 0;
// Store elements to compare
// max_length with its size
// and change the value of
// max_length accordingly
static Vector store = new Vector();
// Store the elements of the
// longest subsequence
static Vector ans = new Vector();
// Function to find the length
// of longest subsequence
static void find_max_length(
int []arr,
int index, int sum, int k)
{
sum = sum + arr[index];
store.add(arr[index]);
if (sum == k)
{
if (max_length < store.size())
{
// Update max_length
max_length = store.size();
// Store the subsequence
// elements
ans = store;
}
}
for (int i = index + 1;
i < arr.length; i++)
{
if (sum + arr[i] <= k)
{
// Recursively proceed
// with obtained sum
find_max_length(arr, i,
sum, k);
// poping elements
// from back
// of vector store
store.remove(store.size() - 1);
}
// if sum > 0 then we don't
// required thatsubsequence
// so return and continue
// with earlier elements
else
return;
}
return;
}
static int longestSubsequence(int []arr,
int n, int k)
{
// Sort the given array
Arrays.sort(arr);
// Traverse the array
for (int i = 0; i < n; i++)
{
// If max_length is already
// greater than or equal
// than remaining length
if (max_length >= n - i)
break;
store.clear();
find_max_length(arr, i, 0, k);
}
return max_length;
}
// Driver code
public static void main(String[] args)
{
int []arr = { -3, 0, 1, 1, 2 };
int n = arr.length;
int k = 1;
System.out.print(longestSubsequence(arr,
n, k));
}
}
// This code is contributed by Princi Singh Python3
# Python3 Program to implement the
# above approach
# Initialise maximum possible
# length of subsequence
max_length = 0
# Store elements to compare
# max_length with its size
# and change the value of
# max_length accordingly
store = []
# Store the elements of the
# longest subsequence
ans = []
# Function to find the length
# of longest subsequence
def find_max_length(arr, index, sum, k):
global max_length
sum = sum + arr[index]
store.append(arr[index])
if (sum == k):
if (max_length < len(store)):
# Update max_length
max_length = len(store)
# Store the subsequence
# elements
ans = store
for i in range ( index + 1, len(arr)):
if (sum + arr[i] <= k):
# Recursively proceed
# with obtained sum
find_max_length(arr, i,
sum, k)
# poping elements
# from back
# of vector store
store.pop()
# if sum > 0 then we don't
# required thatsubsequence
# so return and continue
# with earlier elements
else:
return
return
def longestSubsequence(arr, n, k):
# Sort the given array
arr.sort()
# Traverse the array
for i in range (n):
# If max_length is already
# greater than or equal
# than remaining length
if (max_length >= n - i):
break
store.clear()
find_max_length(arr, i, 0, k)
return max_length
# Driver code
if __name__ == "__main__":
arr = [-3, 0, 1, 1, 2]
n = len(arr)
k = 1
print (longestSubsequence(arr, n, k))
# This code is contributed by ChitranayalC#
// C# program to implement the
// above approach
using System;
using System.Collections.Generic;
class GFG{
// Initialise maximum possible
// length of subsequence
static int max_length = 0;
// Store elements to compare
// max_length with its size
// and change the value of
// max_length accordingly
static List store = new List();
// Store the elements of the
// longest subsequence
static List ans = new List();
// Function to find the length
// of longest subsequence
static void find_max_length(int []arr,
int index,
int sum, int k)
{
sum = sum + arr[index];
store.Add(arr[index]);
if (sum == k)
{
if (max_length < store.Count)
{
// Update max_length
max_length = store.Count;
// Store the subsequence
// elements
ans = store;
}
}
for(int i = index + 1;
i < arr.Length; i++)
{
if (sum + arr[i] <= k)
{
// Recursively proceed
// with obtained sum
find_max_length(arr, i,
sum, k);
// poping elements
// from back
// of vector store
store.RemoveAt(store.Count - 1);
}
// If sum > 0 then we don't
// required thatsubsequence
// so return and continue
// with earlier elements
else
return;
}
return;
}
static int longestSubsequence(int []arr,
int n, int k)
{
// Sort the given array
Array.Sort(arr);
// Traverse the array
for(int i = 0; i < n; i++)
{
// If max_length is already
// greater than or equal
// than remaining length
if (max_length >= n - i)
break;
store.Clear();
find_max_length(arr, i, 0, k);
}
return max_length;
}
// Driver code
public static void Main(String[] args)
{
int []arr = { -3, 0, 1, 1, 2 };
int n = arr.Length;
int k = 1;
Console.Write(longestSubsequence(arr,
n, k));
}
}
// This code is contributed by gauravrajput1 输出：
5
时间复杂度： O(N 3 )
辅助空间： O(N)
动态编程方法：有关解决此问题的进一步优化方法，请参阅本文。