给定一个包含正整数和负整数的长度为N的列表。任务是选择给定序列的最长交替子序列(即每个下一个元素的符号与当前元素的符号相反)。在所有这些子序列中，我们必须选择一个具有最大元素总和的子序列并显示该总和。
例子：
Input: list = [-2 10 3 -8 -4 -1 5 -2 -3 1]
Output: 11
Explaination:
The largest subsequence with the greatest sum is [-2 10 -1 5 -2 1] with length 6.
Input: list=[12 4 -5 7 -9]
Output: 5
Explaination:
The largest subsequence with greatest sum is [12 -5 7 -9] with length 4.
方法：可以通过以下方法达到解决方案：-
- 为了获得具有最大长度和最大和的交替子序列，我们将遍历整个列表(列表长度)-1 次以比较连续元素的符号。
- 在遍历过程中，如果我们得到超过 1 个相同符号的连续元素(exp. 1 2 4)，那么我们会将其中的最大元素附加到另一个名为large 的列表中。所以从 1 2 和 4 我们将 4 附加到另一个列表中。
- 如果我们有连续的相反符号元素，我们只需将这些元素添加到名为large 的列表中。
- 最后，名为large的列表将具有最长的交替子序列和最大的元素。
- 现在，我们必须计算名为large 的列表中所有元素的总和。
Lets take an example, we have a list [1, 2, 3, -2, -5, 1, -7, -1].
- In traversing this list length-1 times, we are getting 1, 2, 3 with the same sign so we will append greatest of these (i.e 3) to another list named large here.
Hence large=[3] - Now -2 and -5 have the same sign so we will append -2 to another List.
large=[3, -2] - Now, the sign of 1 and -7 are opposite, so we will append 1 to large.
large=[3, -2, 1] - For -7, -1 signs are same, Hence append -1 to large.
large=[3, -2, 1, -1] - Calculate the sum = 3 – 2 + 1 – 1 = 1
下面是上述方法的实现：
C++
// C++ implementation to find the
// longest alternating subsequence
// which has the maximum sum
#include
using namespace std;
int calculateMaxSum(int n, int li[])
{
// Creating a temporary list ar to
// every time store same sign element
// to calculate maximum element from
// that list ar
vector ar;
// Appending 1st element of list li
// to the ar
ar.push_back(li[0]);
// Creating list to store maximum
// values
vector large;
for(int j = 0; j < n - 1; j++)
{
// If both number are positive
// then append (j + 1)th element
// to temporary list ar
if(li[j] > 0 and li[j + 1] > 0)
{
ar.push_back(li[j + 1]);
}
else if(li[j] > 0 and li[j + 1] < 0)
{
// If opposite elements found
// then append maximum element
// to large list
large.push_back(*max_element(ar.begin(),
ar.end()));
// Empty ar list to re-append
// next elements
ar.clear();
ar.push_back(li[j + 1]);
}
else if(li[j] < 0 and li[j + 1] > 0)
{
// If opposite elements found
// then append maximum element
// to large list
large.push_back(*max_element(ar.begin(),
ar.end()));
// Empty ar list to re-append
// next elements
ar.clear();
ar.push_back(li[j + 1]);
}
else
{
// If both number are negative
// then append (j + 1)th element
// to temporary list ar
ar.push_back(li[j + 1]);
}
}
// The final Maximum element in ar list
// also needs to be appended to large list
large.push_back(*max_element(ar.begin(),
ar.end()));
// Returning the sum of all elements
// from largest elements list with
// largest alternating subsequence size
int sum = 0;
for(int i = 0; i < large.size(); i++)
sum += large[i];
return sum;
}
// Driver code
int main()
{
int list[] = { -2, 8, 3, 8, -4, -15,
5, -2, -3, 1 };
int N = sizeof(list) / sizeof(list[0]);
cout << (calculateMaxSum(N, list));
}
// This code is contributed by Bhupendra_Singh Java
// Java implementation to find the
// longest alternating subsequence
// which has the maximum sum
import java.util.*;
class GFG{
static int calculateMaxSum(int n, int li[])
{
// Creating a temporary list ar to
// every time store same sign element
// to calculate maximum element from
// that list ar
Vector ar = new Vector<>();
// Appending 1st element of list li
// to the ar
ar.add(li[0]);
// Creating list to store maximum
// values
Vector large = new Vector<>();
for(int j = 0; j < n - 1; j++)
{
// If both number are positive
// then append (j + 1)th element
// to temporary list ar
if(li[j] > 0 && li[j + 1] > 0)
{
ar.add(li[j + 1]);
}
else if(li[j] > 0 && li[j + 1] < 0)
{
// If opposite elements found
// then append maximum element
// to large list
large.add(Collections.max(ar));
// Empty ar list to re-append
// next elements
ar.clear();
ar.add(li[j + 1]);
}
else if(li[j] < 0 && li[j + 1] > 0)
{
// If opposite elements found
// then append maximum element
// to large list
large.add(Collections.max(ar));
// Empty ar list to re-append
// next elements
ar.clear();
ar.add(li[j + 1]);
}
else
{
// If both number are negative
// then append (j + 1)th element
// to temporary list ar
ar.add(li[j + 1]);
}
}
// The final Maximum element in ar list
// also needs to be appended to large list
large.add(Collections.max(ar));
// Returning the sum of all elements
// from largest elements list with
// largest alternating subsequence size
int sum = 0;
for(int i = 0; i < large.size(); i++)
sum += (int)large.get(i);
return sum;
}
// Driver code
public static void main(String args[])
{
int list[] = { -2, 8, 3, 8, -4, -15,
5, -2, -3, 1 };
int N = (list.length);
System.out.print(calculateMaxSum(N, list));
}
}
// This code is contributed by Stream_Cipher Python3
# Python3 implementation to find the
# longest alternating subsequence
# which has the maximum sum
def calculateMaxSum(n, li):
# Creating a temporary list ar to every
# time store same sign element to
# calculate maximum element from
# that list ar
ar =[]
# Appending 1st element of list li
# to the ar
ar.append(li[0])
# Creating list to store maximum
# values
large =[]
for j in range(0, n-1):
# If both number are positive
# then append (j + 1)th element
# to temporary list ar
if(li[j]>0 and li[j + 1]>0):
ar.append(li[j + 1])
elif(li[j]>0 and li[j + 1]<0):
# If opposite elements found
# then append maximum element
# to large list
large.append(max(ar))
# Empty ar list to re-append
# next elements
ar =[]
ar.append(li[j + 1])
elif(li[j]<0 and li[j + 1]>0):
# If opposite elements found
# then append maximum element
# to large list
large.append(max(ar))
# Empty ar list to re-append
# next elements
ar =[]
ar.append(li[j + 1])
else:
# If both number are negative
# then append (j + 1)th element
# to temporary list ar
ar.append(li[j + 1])
# The final Maximum element in ar list
# also needs to be appended to large list
large.append(max(ar))
# returning the sum of all elements
# from largest elements list with
# largest alternating subsequence size
return sum(large)
# Driver code
list =[-2, 8, 3, 8, -4, -15, 5, -2, -3, 1]
N = len(list)
print(calculateMaxSum(N, list))C#
// C# implementation to find the
// longest alternating subsequence
// which has the maximum sum
using System;
using System.Collections.Generic;
class GFG{
static int find_max(List ar)
{
int mx = -1000000;
foreach(var i in ar)
{
if(i > mx)
mx = i;
}
return mx;
}
static int calculateMaxSum(int n, int []li)
{
// Creating a temporary list ar to
// every time store same sign element
// to calculate maximum element from
// that list ar
List ar = new List();
// Appending 1st element of list li
// to the ar
ar.Add(li[0]);
// Creating list to store maximum
// values
List large = new List();
for(int j = 0; j < n - 1; j++)
{
// If both number are positive
// then append (j + 1)th element
// to temporary list ar
if(li[j] > 0 && li[j + 1] > 0)
{
ar.Add(li[j + 1]);
}
else if(li[j] > 0 && li[j + 1] < 0)
{
// If opposite elements found
// then append maximum element
// to large list
large.Add(find_max(ar));
// Empty ar list to re-append
// next elements
ar.Clear();
ar.Add(li[j + 1]);
}
else if(li[j] < 0 && li[j + 1] > 0)
{
// If opposite elements found
// then append maximum element
// to large list
large.Add(find_max(ar));
// Empty ar list to re-append
// next elements
ar.Clear();
ar.Add(li[j + 1]);
}
else
{
// If both number are negative
// then append (j + 1)th element
// to temporary list ar
ar.Add(li[j + 1]);
}
}
// The final Maximum element in ar list
// also needs to be appended to large list
large.Add(find_max(ar));
// Returning the sum of all elements
// from largest elements list with
// largest alternating subsequence size
int sum = 0;
foreach(var i in large)
{
sum += i;
}
return sum;
}
// Driver code
public static void Main()
{
int []list = { -2, 8, 3, 8, -4, -15,
5, -2, -3, 1 };
int N = (list.Length);
Console.WriteLine(calculateMaxSum(N, list));
}
}
// This code is contributed by Stream_Cipher 输出：
6
时间复杂度： O(N)
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