找出数组第一和第二部分的最大元素
给定一个由 N 个整数组成的数组 arr[] 。任务是在数组的前半部分和后半部分找到最大的元素。注意如果数组的大小是奇数，那么中间的元素将包含在两半中。 示例:
输入: arr[] = {1，12，14，5} 输出: 12，14 前半部分为{1，12}，后半部分为{14，5}。 输入: arr[] = {1，2，3，4，5} 输出: 3，5
方法:计算数组的中间索引为 mid = N / 2 。现在，如果 N 为偶数，第一半元素将出现在子阵列arr[0…中间-1] 和arr[中间…N-1] 中。 如果 N 为奇数，则两半为arr【0…中间】和 arr【中间…N-1】 以下是上述方法的实现:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to print largest element in
// first half and second half of an array
void findMax(int arr[], int n)
{
// To store the maximum element
// in the first half
int maxFirst = INT_MIN;
// Middle index of the array
int mid = n / 2;
// Calculate the maximum element
// in the first half
for (int i = 0; i < mid; i++)
maxFirst = max(maxFirst, arr[i]);
// If the size of array is odd then
// the middle element will be included
// in both the halves
if (n % 2 == 1)
maxFirst = max(maxFirst, arr[mid]);
// To store the maximum element
// in the second half
int maxSecond = INT_MIN;
// Calculate the maximum element
// int the second half
for (int i = mid; i < n; i++)
maxSecond = max(maxSecond, arr[i]);
// Print the found maximums
cout << maxFirst << ", " << maxSecond;
}
// Driver code
int main()
{
int arr[] = { 1, 12, 14, 5 };
int n = sizeof(arr) / sizeof(arr[0]);
findMax(arr, n);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the approach
import java.io.*;
class GFG
{
static void findMax(int []arr, int n)
{
// To store the maximum element
// in the first half
int maxFirst = Integer.MIN_VALUE;
// Middle index of the array
int mid = n / 2;
// Calculate the maximum element
// in the first half
for (int i = 0; i < mid; i++)
{
maxFirst = Math.max(maxFirst, arr[i]);
}
// If the size of array is odd then
// the middle element will be included
// in both the halves
if (n % 2 == 1)
{
maxFirst = Math.max(maxFirst, arr[mid]);
}
// To store the maximum element
// in the second half
int maxSecond = Integer.MIN_VALUE;
// Calculate the maximum element
// int the second half
for (int i = mid; i < n; i++)
{
maxSecond = Math.max(maxSecond, arr[i]);
}
// Print the found maximums
System.out.print(maxFirst + ", " + maxSecond);
// cout << maxFirst << ", " << maxSecond;
}
// Driver Code
public static void main(String[] args)
{
int []arr = { 1, 12, 14, 5 };
int n = arr.length;
findMax(arr, n);
}
}
// This code is contributed by anuj_67..
Python 3
# Python3 implementation of the approach
import sys
# Function to print largest element in
# first half and second half of an array
def findMax(arr, n) :
# To store the maximum element
# in the first half
maxFirst = -sys.maxsize - 1
# Middle index of the array
mid = n // 2;
# Calculate the maximum element
# in the first half
for i in range(0, mid):
maxFirst = max(maxFirst, arr[i])
# If the size of array is odd then
# the middle element will be included
# in both the halves
if (n % 2 == 1):
maxFirst = max(maxFirst, arr[mid])
# To store the maximum element
# in the second half
maxSecond = -sys.maxsize - 1
# Calculate the maximum element
# int the second half
for i in range(mid, n):
maxSecond = max(maxSecond, arr[i])
# Print the found maximums
print(maxFirst, ",", maxSecond)
# Driver code
arr = [1, 12, 14, 5 ]
n = len(arr)
findMax(arr, n)
# This code is contributed by ihritik
C
// C# implementation of the approach
using System;
class GFG
{
static void findMax(int []arr, int n)
{
// To store the maximum element
// in the first half
int maxFirst = int.MinValue;
// Middle index of the array
int mid = n / 2;
// Calculate the maximum element
// in the first half
for (int i = 0; i < mid; i++)
{
maxFirst = Math.Max(maxFirst, arr[i]);
}
// If the size of array is odd then
// the middle element will be included
// in both the halves
if (n % 2 == 1)
{
maxFirst = Math.Max(maxFirst, arr[mid]);
}
// To store the maximum element
// in the second half
int maxSecond = int.MinValue;
// Calculate the maximum element
// int the second half
for (int i = mid; i < n; i++)
{
maxSecond = Math.Max(maxSecond, arr[i]);
}
// Print the found maximums
Console.WriteLine(maxFirst + ", " + maxSecond);
// cout << maxFirst << ", " << maxSecond;
}
// Driver Code
public static void Main()
{
int []arr = { 1, 12, 14, 5 };
int n = arr.Length;
findMax(arr, n);
}
}
// This code is contributed by nidhiva
java 描述语言
// javascript implementation of the approach
function findMax(arr, n)
{
// To store the maximum element
// in the first half
var maxFirst = Number.MIN_VALUE
// Middle index of the array
var mid = n / 2;
// Calculate the maximum element
// in the first half
for (var i = 0; i < mid; i++)
{
maxFirst = Math.max(maxFirst, arr[i]);
}
// If the size of array is odd then
// the middle element will be included
// in both the halves
if (n % 2 == 1)
{
maxFirst = Math.max(maxFirst, arr[mid]);
}
// To store the maximum element
// in the second half
var maxSecond = Number.MIN_VALUE
// Calculate the maximum element
// int the second half
for (var i = mid; i < n; i++)
{
maxSecond = Math.max(maxSecond, arr[i]);
}
// Print the found maximums
document.write(maxFirst + ", " + maxSecond);
}
// Driver Code
var arr = [ 1, 12, 14, 5 ];
var n = arr.length;
findMax(arr, n);
// This code is contributed by bunnyram19.
Output:
12, 14
时间复杂度: O(n)
辅助空间: O(1)
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