数组中非相邻元素的最大求和对
给定一个不同整数的数组arr[] ，找到最大和的对，使得该对的两个元素在数组中不相邻
例子：
Input: arr[] = {7, 3, 4, 1, 8, 9}
Output: 9 7
Explanation: Pair with maximum sum is (8, 9).
But since both elements are adjacent, it is not a valid pair.
Therefore, the pair with maximum sum is (9, 7) with sum 16.
Input: arr[] = {2, 1, 3}
Output: 2 3
Input: arr[] = {4, 3, 7, 9, 8, 1}
Output: 7 8
Explanation: Sum of both the pairs {7, 9} and {9, 8} are greater.
But they are adjacent pairs. Hence this is the valid answer.
方法：想法是计算数组中最大的三个元素的索引。计算索引后，检查具有最大总和的对，并检查它们是否相邻。由于已经计算了三个最大的元素，因此总是可以得到至少一对不相邻的两个元素。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to find the max pair sum
// which are non-adjacent
pair getMaxSumPair(int arr[], int N)
{
// Indices for storing first, second
// and third largest element
if (N < 3) {
return make_pair(-1, -1);
}
int first, second, third;
if (arr[1] > arr[0]) {
if (arr[1] > arr[2]) {
first = 1;
if (arr[2] > arr[0]) {
second = 2;
third = 0;
}
else {
second = 0;
third = 2;
}
}
else {
first = 2;
second = 1;
third = 0;
}
}
else {
if (arr[0] > arr[2]) {
first = 0;
if (arr[2] > arr[1]) {
second = 2;
third = 1;
}
else {
second = 1;
third = 2;
}
}
else {
first = 2;
second = 0;
third = 1;
}
}
for (int i = 3; i < N; ++i) {
if (arr[i] > arr[first]) {
third = second;
second = first;
first = i;
}
else if (arr[i] > arr[second]) {
third = second;
second = i;
}
else if (arr[i] > arr[third]) {
third = i;
}
}
// Check whether the elements
// are adjacent or not
if (abs(first - second) == 1) {
if (abs(first - third) == 1) {
return make_pair(arr[second],
arr[third]);
}
else {
return make_pair(arr[first],
arr[third]);
}
}
else {
return make_pair(arr[first],
arr[second]);
}
}
// Driver Code
int main()
{
int arr[] = { 7, 3, 4, 1, 8, 9 };
int N = sizeof(arr) / sizeof(*arr);
pair p = getMaxSumPair(arr, N);
cout << p.first << " " << p.second;
return 0;
} Java
// Java program for the above approach
class GFG {
static class Pair {
int first;
int second;
public Pair(int first, int second) {
this.first = first;
this.second = second;
}
}
// Function to find the max pair sum
// which are non-adjacent
static Pair getMaxSumPair(int arr[], int N)
{
// Indices for storing first, second
// and third largest element
if (N < 3) {
return new Pair(-1, -1);
}
int first, second, third;
if (arr[1] > arr[0]) {
if (arr[1] > arr[2]) {
first = 1;
if (arr[2] > arr[0]) {
second = 2;
third = 0;
} else {
second = 0;
third = 2;
}
} else {
first = 2;
second = 1;
third = 0;
}
} else {
if (arr[0] > arr[2]) {
first = 0;
if (arr[2] > arr[1]) {
second = 2;
third = 1;
} else {
second = 1;
third = 2;
}
} else {
first = 2;
second = 0;
third = 1;
}
}
for (int i = 3; i < N; ++i) {
if (arr[i] > arr[first]) {
third = second;
second = first;
first = i;
} else if (arr[i] > arr[second]) {
third = second;
second = i;
} else if (arr[i] > arr[third]) {
third = i;
}
}
// Check whether the elements
// are adjacent or not
if (Math.abs(first - second) == 1) {
if (Math.abs(first - third) == 1) {
return new Pair(arr[second], arr[third]);
} else {
return new Pair(arr[first], arr[third]);
}
} else {
return new Pair(arr[first], arr[second]);
}
}
// Driver Code
public static void main(String args[]) {
int arr[] = { 7, 3, 4, 1, 8, 9 };
int N = arr.length;
Pair p = getMaxSumPair(arr, N);
System.out.println(p.first + " " + p.second);
}
}
// This code is contributed by saurabh_jaiswal.Python3
# Python code for the above approach
# Function to find the max pair sum
# which are non-adjacent
def getMaxSumPair(arr, N):
# Indices for storing first, second
# and third largest element
if (N < 3):
return [-1, -1]
first = second = third = None
if (arr[1] > arr[0]):
if (arr[1] > arr[2]):
first = 1
if (arr[2] > arr[0]):
second = 2
third = 0
else:
second = 0
third = 2
else:
first = 2
second = 1
third = 0
else:
if (arr[0] > arr[2]):
first = 0
if (arr[2] > arr[1]):
second = 2
third = 1
else:
second = 1
third = 2
else:
first = 2
second = 0
third = 1
for i in range(3, N):
if (arr[i] > arr[first]):
third = second
second = first
first = i
elif (arr[i] > arr[second]):
third = second
second = i
elif (arr[i] > arr[third]):
third = i
# Check whether the elements
# are adjacent or not
if (abs(first - second) == 1):
if (abs(first - third) == 1):
return [arr[second],
arr[third]]
else:
return [arr[first],
arr[third]]
else:
return [arr[first],
arr[second]]
# Driver Code
arr = [7, 3, 4, 1, 8, 9]
N = len(arr)
p = getMaxSumPair(arr, N)
print(f"{p[0]} {p[1]}")
# This code is contributed by Saurabh JaiswalC#
// C# program for the above approach
using System;
public class GFG {
class Pair {
public int first;
public int second;
public Pair(int first, int second) {
this.first = first;
this.second = second;
}
}
// Function to find the max pair sum
// which are non-adjacent
static Pair getMaxSumPair(int []arr, int N)
{
// Indices for storing first, second
// and third largest element
if (N < 3) {
return new Pair(-1, -1);
}
int first, second, third;
if (arr[1] > arr[0]) {
if (arr[1] > arr[2]) {
first = 1;
if (arr[2] > arr[0]) {
second = 2;
third = 0;
} else {
second = 0;
third = 2;
}
} else {
first = 2;
second = 1;
third = 0;
}
} else {
if (arr[0] > arr[2]) {
first = 0;
if (arr[2] > arr[1]) {
second = 2;
third = 1;
} else {
second = 1;
third = 2;
}
} else {
first = 2;
second = 0;
third = 1;
}
}
for (int i = 3; i < N; ++i) {
if (arr[i] > arr[first]) {
third = second;
second = first;
first = i;
} else if (arr[i] > arr[second]) {
third = second;
second = i;
} else if (arr[i] > arr[third]) {
third = i;
}
}
// Check whether the elements
// are adjacent or not
if (Math.Abs(first - second) == 1) {
if (Math.Abs(first - third) == 1) {
return new Pair(arr[second], arr[third]);
} else {
return new Pair(arr[first], arr[third]);
}
} else {
return new Pair(arr[first], arr[second]);
}
}
// Driver Code
public static void Main(String []args) {
int []arr = { 7, 3, 4, 1, 8, 9 };
int N = arr.Length;
Pair p = getMaxSumPair(arr, N);
Console.WriteLine(p.first + " " + p.second);
}
}
// This code is contributed by shikhasingrajputJavascript
输出
9 7时间复杂度：O(N)
辅助空间：O(1)