给定大小为n的数组arr [] ，其中有正整数和负整数(不包括零)。任务是找到具有最大大小和最大和的交替符号的子序列，即在每个相邻元素的子序列符号中是相反的，例如，如果第一个为正，则第二个必须为负，然后为另一个正整数等等。
例子：
Input: arr[] = {2, 3, 7, -6, -4}
Output: 7 -4
Explanation:
Possible subsequences are [2, -6] [2, -4] [3, -6] [3, -4] [7, -6] [7, -4].
Out of these [7, -4] has the maximum sum.
Input: arr[] = {-4, 9, 4, 11, -5, -17, 9, -3, -5, 2}
Output: -4 11 -5 9 -3 2
方法：
解决上述问题的主要思想是从包含相同符号的数组段中找到最大元素，这意味着我们必须从连续的正负元素中选择最大的元素。由于我们需要最大大小，因此我们将仅从每个段中获取一个元素，并且为了使总和最大化，我们需要获取每个段中的最大元素。
下面是上述方法的实现：
CPP
// C++ implementation to find the
// subsequence with alternating sign
// having maximum size and maximum sum.
#include
using namespace std;
// Function to find the subsequence
// with alternating sign having
// maximum size and maximum sum.
void findSubsequence(int arr[], int n)
{
int sign[n] = { 0 };
// Find whether each element
// is positive or negative
for (int i = 0; i < n; i++) {
if (arr[i] > 0)
sign[i] = 1;
else
sign[i] = -1;
}
int k = 0;
int result[n] = { 0 };
// Find the required subsequence
for (int i = 0; i < n; i++) {
int cur = arr[i];
int j = i;
while (j < n && sign[i] == sign[j]) {
// Find the maximum element
// in the specified range
cur = max(cur, arr[j]);
++j;
}
result[k++] = cur;
i = j - 1;
}
// print the result
for (int i = 0; i < k; i++)
cout << result[i] << " ";
cout << "\n";
}
// Driver code
int main()
{
// array declaration
int arr[] = { -4, 9, 4, 11, -5, -17, 9, -3, -5, 2 };
// size of array
int n = sizeof(arr) / sizeof(arr[0]);
findSubsequence(arr, n);
return 0;
} Java
// Java implementation to find the
// subsequence with alternating sign
// having maximum size and maximum sum.
class GFG{
// Function to find the subsequence
// with alternating sign having
// maximum size and maximum sum.
static void findSubsequence(int arr[], int n)
{
int sign[] = new int[n];
// Find whether each element
// is positive or negative
for (int i = 0; i < n; i++) {
if (arr[i] > 0)
sign[i] = 1;
else
sign[i] = -1;
}
int k = 0;
int result[] = new int[n];
// Find the required subsequence
for (int i = 0; i < n; i++) {
int cur = arr[i];
int j = i;
while (j < n && sign[i] == sign[j]) {
// Find the maximum element
// in the specified range
cur = Math.max(cur, arr[j]);
++j;
}
result[k++] = cur;
i = j - 1;
}
// print the result
for (int i = 0; i < k; i++)
System.out.print(result[i]+ " ");
System.out.print("\n");
}
// Driver code
public static void main(String[] args)
{
// array declaration
int arr[] = { -4, 9, 4, 11, -5, -17, 9, -3, -5, 2 };
// size of array
int n = arr.length;
findSubsequence(arr, n);
}
}
// This code is contributed by Princi SinghPython3
# Python3 implementation to find the
# subsequence with alternating sign
# having maximum size and maximum sum.
# Function to find the subsequence
# with alternating sign having
# maximum size and maximum sum.
def findSubsequence(arr, n):
sign = [0]*n
# Find whether each element
# is positive or negative
for i in range(n):
if (arr[i] > 0):
sign[i] = 1
else:
sign[i] = -1
k = 0
result = [0]*n
# Find the required subsequence
i = 0
while i < n:
cur = arr[i]
j = i
while (j < n and sign[i] == sign[j]):
# Find the maximum element
# in the specified range
cur = max(cur, arr[j])
j += 1
result[k] = cur
k += 1
i = j - 1
i += 1
# print the result
for i in range(k):
print(result[i],end=" ")
# Driver code
if __name__ == '__main__':
# array declaration
arr=[-4, 9, 4, 11, -5, -17, 9, -3, -5, 2]
# size of array
n = len(arr)
findSubsequence(arr, n)
# This code is contributed by mohit kumar 29C#
// C# implementation to find the
// subsequence with alternating sign
// having maximum size and maximum sum.
using System;
public class GFG{
// Function to find the subsequence
// with alternating sign having
// maximum size and maximum sum.
static void findSubsequence(int []arr, int n)
{
int []sign = new int[n];
// Find whether each element
// is positive or negative
for (int i = 0; i < n; i++) {
if (arr[i] > 0)
sign[i] = 1;
else
sign[i] = -1;
}
int k = 0;
int []result = new int[n];
// Find the required subsequence
for (int i = 0; i < n; i++) {
int cur = arr[i];
int j = i;
while (j < n && sign[i] == sign[j]) {
// Find the maximum element
// in the specified range
cur = Math.Max(cur, arr[j]);
++j;
}
result[k++] = cur;
i = j - 1;
}
// print the result
for (int i = 0; i < k; i++)
Console.Write(result[i]+ " ");
Console.Write("\n");
}
// Driver code
public static void Main(String[] args)
{
// array declaration
int []arr = { -4, 9, 4, 11, -5, -17, 9, -3, -5, 2 };
// size of array
int n = arr.Length;
findSubsequence(arr, n);
}
}
// This code contributed by Rajput-Ji输出：
-4 11 -5 9 -3 2
时间复杂度： O(N)
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