数组中的最大三元组总和
原文： https://www.geeksforgeeks.org/maximum-triplet-sum-array/
给定一个数组，任务是在数组中找到最大三元组和。
示例：
Input : arr[] = {1, 2, 3, 0, -1, 8, 10}
Output : 21
10 + 8 + 3 = 21
Input : arr[] = {9, 8, 20, 3, 4, -1, 0}
Output : 37
20 + 9 + 8 = 37
朴素的方法：在此方法中，我们简单地运行三个循环，然后一个接一个地添加三个元素，如果三个元素的和更大，则与先前的和进行比较，然后存储在先前的和中。
C++
// C++ code to find maximum triplet sum
#include <bits/stdc++.h>
using namespace std;
int maxTripletSum(int arr[], int n)
{
// Initialize sum with INT_MIN
int sum = INT_MIN;
for (int i = 0; i < n; i++)
for (int j = i + 1; j < n; j++)
for (int k = j + 1; k < n; k++)
if (sum < arr[i] + arr[j] + arr[k])
sum = arr[i] + arr[j] + arr[k];
return sum;
}
// Driven code
int main()
{
int arr[] = { 1, 0, 8, 6, 4, 2 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << maxTripletSum(arr, n);
return 0;
}
Java
// Java code to find maximum triplet sum
import java.io.*;
class GFG {
static int maxTripletSum(int arr[], int n)
{
// Initialize sum with INT_MIN
int sum = -1000000;
for (int i = 0; i < n; i++)
for (int j = i + 1; j < n; j++)
for (int k = j + 1; k < n; k++)
if (sum < arr[i] + arr[j] + arr[k])
sum = arr[i] + arr[j] + arr[k];
return sum;
}
// Driven code
public static void main(String args[])
{
int arr[] = { 1, 0, 8, 6, 4, 2 };
int n = arr.length;
System.out.println(maxTripletSum(arr, n));
}
}
// This code is contributed by Nikita Tiwari.
Python3
# Python 3 code to find
# maximum triplet sum
def maxTripletSum(arr, n) :
# Initialize sum with
# INT_MIN
sm = -1000000
for i in range(0, n) :
for j in range(i + 1, n) :
for k in range(j + 1, n) :
if (sm < (arr[i] + arr[j] + arr[k])) :
sm = arr[i] + arr[j] + arr[k]
return sm
# Driven code
arr = [ 1, 0, 8, 6, 4, 2 ]
n = len(arr)
print(maxTripletSum(arr, n))
# This code is contributed by Nikita Tiwari.
C#
// C# code to find maximum triplet sum
using System;
class GFG {
static int maxTripletSum(int[] arr, int n)
{
// Initialize sum with INT_MIN
int sum = -1000000;
for (int i = 0; i < n; i++)
for (int j = i + 1; j < n; j++)
for (int k = j + 1; k < n; k++)
if (sum < arr[i] + arr[j] + arr[k])
sum = arr[i] + arr[j] + arr[k];
return sum;
}
// Driven code
public static void Main()
{
int[] arr = { 1, 0, 8, 6, 4, 2 };
int n = arr.Length;
Console.WriteLine(maxTripletSum(arr, n));
}
}
// This code is contributed by vt_m.
PHP
<?php
// PHP code to find maximum triplet sum
function maxTripletSum( $arr, $n)
{
// Initialize sum with INT_MIN
$sum = PHP_INT_MIN;
for($i = 0; $i < $n; $i++)
for($j = $i + 1; $j < $n; $j++)
for($k = $j + 1; $k < $n; $k++)
if ($sum < $arr[$i] +
$arr[$j] +
$arr[$k])
$sum = $arr[$i] +
$arr[$j] +
$arr[$k];
return $sum;
}
// Driver Code
$arr = array(1, 0, 8, 6, 4, 2);
$n = count($arr);
echo maxTripletSum($arr, $n);
// This code is contributed by anuj_67\.
?>
输出：
18
时间复杂度：O(n ^ 3)。
空间复杂度：O(1)。
另一种方法：在这种情况下，我们首先需要对整个数组进行排序，然后，当我们添加数组的最后三个元素时，我们将找到三叉树的最大和。
C++
// C++ code to find maximum triplet sum
#include <bits/stdc++.h>
using namespace std;
// This function assumes that there are at least
// three elements in arr[].
int maxTripletSum(int arr[], int n)
{
// sort the given array
sort(arr, arr + n);
// After sorting the array.
// Add last three element of the given array
return arr[n - 1] + arr[n - 2] + arr[n - 3];
}
// Driven code
int main()
{
int arr[] = { 1, 0, 8, 6, 4, 2 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << maxTripletSum(arr, n);
return 0;
}
Java
// Java code to find maximum triplet sum
import java.io.*;
import java.util.*;
class GFG {
// This function assumes that there are
// at least three elements in arr[].
static int maxTripletSum(int arr[], int n)
{
// sort the given array
Arrays.sort(arr);
// After sorting the array.
// Add last three element
// of the given array
return arr[n - 1] + arr[n - 2] + arr[n - 3];
}
// Driven code
public static void main(String args[])
{
int arr[] = { 1, 0, 8, 6, 4, 2 };
int n = arr.length;
System.out.println(maxTripletSum(arr, n));
}
}
// This code is contributed by Nikita Tiwari.
Python3
# Python 3 code to find
# maximum triplet sum
# This function assumes
# that there are at least
# three elements in arr[].
def maxTripletSum(arr, n) :
# sort the given array
arr.sort()
# After sorting the array.
# Add last three element
# of the given array
return (arr[n - 1] + arr[n - 2] + arr[n - 3])
# Driven code
arr = [ 1, 0, 8, 6, 4, 2 ]
n = len(arr)
print(maxTripletSum(arr, n))
# This code is contributed by Nikita Tiwari.
C
// C# code to find maximum triplet sum
using System;
class GFG {
// This function assumes that there are
// at least three elements in arr[].
static int maxTripletSum(int[] arr, int n)
{
// sort the given array
Array.Sort(arr);
// After sorting the array.
// Add last three element
// of the given array
return arr[n - 1] + arr[n - 2] + arr[n - 3];
}
// Driven code
public static void Main()
{
int[] arr = { 1, 0, 8, 6, 4, 2 };
int n = arr.Length;
Console.WriteLine(maxTripletSum(arr, n));
}
}
// This code is contributed by vt_m.
PHP
<?php
// PHP code to find
// maximum triplet sum
// This function assumes that
// there are at least
// three elements in arr[].
function maxTripletSum( $arr, $n)
{
// sort the given array
sort($arr);
// After sorting the array.
// Add last three element
// of the given array
return $arr[$n - 1] + $arr[$n - 2] +
$arr[$n - 3];
}
// Driver code
$arr = array( 1, 0, 8, 6, 4, 2 );
$n = count($arr);
echo maxTripletSum($arr, $n);
// This code is contributed by anuj_67\.
?>
输出：
18
时间复杂度：O(nLogn)。
空间复杂度：O(1)。
有效方法：扫描数组并计算数组中存在的最大值，第二大和第三最大元素，并返回其总和，即为最大和。
C++
// C++ code to find maximum triplet sum
#include <bits/stdc++.h>
using namespace std;
// This function assumes that there are at least
// three elements in arr[].
int maxTripletSum(int arr[], int n)
{
// Initialize Maximum, second maximum and third
// maximum element
int maxA = INT_MIN, maxB = INT_MIN, maxC = INT_MIN;
for (int i = 0; i < n; i++) {
// Update Maximum, second maximum and third
// maximum element
if (arr[i] > maxA) {
maxC = maxB;
maxB = maxA;
maxA = arr[i];
}
// Update second maximum and third maximum
// element
else if (arr[i] > maxB) {
maxC = maxB;
maxB = arr[i];
}
// Update third maximum element
else if (arr[i] > maxC)
maxC = arr[i];
}
return (maxA + maxB + maxC);
}
// Driven code
int main()
{
int arr[] = { 1, 0, 8, 6, 4, 2 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << maxTripletSum(arr, n);
return 0;
}
Java
// Java code to find maximum triplet sum
import java.io.*;
import java.util.*;
class GFG {
// This function assumes that there
// are at least three elements in arr[].
static int maxTripletSum(int arr[], int n)
{
// Initialize Maximum, second maximum and third
// maximum element
int maxA = -100000000, maxB = -100000000;
int maxC = -100000000;
for (int i = 0; i < n; i++) {
// Update Maximum, second maximum
// and third maximum element
if (arr[i] > maxA)
{
maxC = maxB;
maxB = maxA;
maxA = arr[i];
}
// Update second maximum and third maximum
// element
else if (arr[i] > maxB)
{
maxC = maxB;
maxB = arr[i];
}
// Update third maximum element
else if (arr[i] > maxC)
maxC = arr[i];
}
return (maxA + maxB + maxC);
}
// Driven code
public static void main(String args[])
{
int arr[] = { 1, 0, 8, 6, 4, 2 };
int n = arr.length;
System.out.println(maxTripletSum(arr, n));
}
}
// This code is contributed by Nikita Tiwari.
Python3
# Python 3 code to find
# maximum triplet sum
# This function assumes
# that there are at least
# three elements in arr[].
def maxTripletSum(arr, n) :
# Initialize Maximum, second
# maximum and third maximum
# element
maxA = -100000000
maxB = -100000000
maxC = -100000000
for i in range(0, n) :
# Update Maximum, second maximum
# and third maximum element
if (arr[i] > maxA) :
maxC = maxB
maxB = maxA
maxA = arr[i]
# Update second maximum and
# third maximum element
elif (arr[i] > maxB) :
maxC = maxB
maxB = arr[i]
# Update third maximum element
elif (arr[i] > maxC) :
maxC = arr[i]
return (maxA + maxB + maxC)
# Driven code
arr = [ 1, 0, 8, 6, 4, 2 ]
n = len(arr)
print(maxTripletSum(arr, n))
# This code is contributed by Nikita Tiwari.
C
// C# code to find maximum triplet sum
using System;
class GFG {
// This function assumes that there
// are at least three elements in arr[].
static int maxTripletSum(int[] arr, int n)
{
// Initialize Maximum, second maximum
// and third maximum element
int maxA = -100000000, maxB = -100000000;
int maxC = -100000000;
for (int i = 0; i < n; i++) {
// Update Maximum, second maximum
// and third maximum element
if (arr[i] > maxA) {
maxC = maxB;
maxB = maxA;
maxA = arr[i];
}
// Update second maximum and third
// maximum element
else if (arr[i] > maxB) {
maxC = maxB;
maxB = arr[i];
}
// Update third maximum element
else if (arr[i] > maxC)
maxC = arr[i];
}
return (maxA + maxB + maxC);
}
// Driven code
public static void Main()
{
int[] arr = { 1, 0, 8, 6, 4, 2 };
int n = arr.Length;
Console.WriteLine(maxTripletSum(arr, n));
}
}
// This code is contributed by vt_m.
PHP
<?php
// PHP code to find
// maximum triplet sum
// This function assumes that
// there are at least three
// elements in arr[].
function maxTripletSum($arr, $n)
{
// Initialize Maximum,
// second maximum and
// third maximum element
$maxA = PHP_INT_MIN;
$maxB = PHP_INT_MIN;
$maxC = PHP_INT_MIN;
for ( $i = 0; $i < $n; $i++)
{
// Update Maximum,
// second maximum and
// third maximum element
if ($arr[$i] > $maxA)
{
$maxC = $maxB;
$maxB = $maxA;
$maxA = $arr[$i];
}
// Update second maximum and
// third maximum element
else if ($arr[$i] > $maxB)
{
$maxC = $maxB;
$maxB = $arr[$i];
}
// Update third maximum element
else if ($arr[$i] > $maxC)
$maxC = $arr[$i];
}
return ($maxA + $maxB + $maxC);
}
// Driven code
$arr = array( 1, 0, 8, 6, 4, 2 );
$n = count($arr);
echo maxTripletSum($arr, $n);
// This code is contributed by anuj_67\.
?>
输出：
18
时间复杂度：O(n)。
空间复杂度：O(1)。
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