给定一个由N 个整数和一个整数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.
朴素方法：解决问题的最简单方法是生成所有不同长度的可能子序列，并检查它们的总和是否等于K。在总和为K的所有这些子序列中，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：这个问题的基本方法是对向量进行排序并找到所有可能子序列的总和，然后选择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，用于存储最大长度子集。
- 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，其中元素在[i, N-1]范围内的总和为K 。
- 每次找到总和为K的子集时，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。
- 在计算所有可能的子集总和后，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 Javascript
输出：
5时间复杂度： O(N 3 )
辅助空间： O(N)
动态规划方法：参考这篇文章，进一步优化解决问题的方法。
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