排序和旋转数组中的最大元素

给定一个不同元素的排序数组arr [] ，该数组在某个未知点旋转，任务是在其中找到最大元素。

例子：

Input: arr[] = {3, 4, 5, 1, 2}
Output: 5

Input: arr[] = {1, 2, 3}
Output: 3

方法：一个简单的解决方案是遍历整个数组并找到最大值。该解决方案需要O(n)时间。
我们可以使用Binary Search在O(Logn)中进行操作。如果我们仔细看一下上面的示例，我们可以很容易地找出以下模式：

• maximum元素是next小于它的唯一元素。如果没有下一个较小的元素，则不会旋转(最后一个元素为最大值)。我们通过与中– 1和中+ 1的元素进行比较来检查此条件是否存在中间元素。
• 如果最大元素不在中间(既不是中也不是中+1)，则最大元素位于左半部分或右半部分。
  1. 如果中间元素大于最后一个元素，则最大元素位于左半部分。
  2. 其他最大元素位于右半部分。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
  
// Function to return the maximum element
int findMax(int arr[], int low, int high)
{
  
    // This condition is for the case when
    // array is not rotated at all
    if (high < low)
        return arr[0];
  
    // If there is only one element left
    if (high == low)
        return arr[low];
  
    // Find mid
    int mid = low + (high - low) / 2;
  
    // Check if mid itself is maximum element
    if (mid < high && arr[mid + 1] < arr[mid]) {
        return arr[mid];
    }
  
    // Check if element at (mid - 1) is maximum element
    // Consider the cases like {4, 5, 1, 2, 3}
    if (mid > low && arr[mid] < arr[mid - 1]) {
        return arr[mid - 1];
    }
  
    // Decide whether we need to go to
    // the left half or the right half
    if (arr[low] > arr[mid]) {
        return findMax(arr, low, mid - 1);
    }
    else {
        return findMax(arr, mid + 1, high);
    }
}
  
// Driver code
int main()
{
    int arr[] = { 5, 6, 1, 2, 3, 4 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << findMax(arr, 0, n - 1);
  
    return 0;
}

Java

// Java implementation of the approach
class GFG
{
      
// Function to return the maximum element
static int findMax(int arr[], int low, int high)
{
  
    // This condition is for the case when
    // array is not rotated at all
    if (high < low)
        return arr[0];
  
    // If there is only one element left
    if (high == low)
        return arr[low];
  
    // Find mid
    int mid = low + (high - low) / 2;
  
    // Check if mid itself is maximum element
    if (mid < high && arr[mid + 1] < arr[mid])
    {
        return arr[mid];
    }
  
    // Check if element at (mid - 1) is maximum element
    // Consider the cases like {4, 5, 1, 2, 3}
    if (mid > low && arr[mid] < arr[mid - 1])
    {
        return arr[mid - 1];
    }
  
    // Decide whether we need to go to
    // the left half or the right half
    if (arr[low] > arr[mid])
    {
        return findMax(arr, low, mid - 1);
    }
    else 
    {
        return findMax(arr, mid + 1, high);
    }
}
  
// Driver code
public static void main(String[] args)
{
    int arr[] = { 5, 6, 1, 2, 3, 4 };
    int n = arr.length;
    System.out.println(findMax(arr, 0, n - 1));
}
}
  
// This code is contributed by Code_Mech.

Python3

# Python3 implementation of the approach
  
# Function to return the maximum element
def findMax(arr, low, high):
  
    # This condition is for the case when
    # array is not rotated at all
    if (high < low):
        return arr[0]
  
    # If there is only one element left
    if (high == low):
        return arr[low]
  
    # Find mid
    mid = low + (high - low) // 2
  
    # Check if mid itself is maximum element
    if (mid < high and arr[mid + 1] < arr[mid]):
        return arr[mid]
      
    # Check if element at (mid - 1) is maximum element
    # Consider the cases like {4, 5, 1, 2, 3}
    if (mid > low and arr[mid] < arr[mid - 1]):
        return arr[mid - 1]
  
    # Decide whether we need to go to
    # the left half or the right half
    if (arr[low] > arr[mid]):
        return findMax(arr, low, mid - 1)
    else:
        return findMax(arr, mid + 1, high)
  
# Driver code
arr = [5, 6, 1, 2, 3, 4]
n = len(arr)
print(findMax(arr, 0, n - 1))
  
# This code is contributed by mohit kumar

C#

// C# implementation of the approach
using System;
  
class GFG
{
      
// Function to return the maximum element
static int findMax(int []arr, 
                   int low, int high)
{
  
    // This condition is for the case 
    // when array is not rotated at all
    if (high < low)
        return arr[0];
  
    // If there is only one element left
    if (high == low)
        return arr[low];
  
    // Find mid
    int mid = low + (high - low) / 2;
  
    // Check if mid itself is maximum element
    if (mid < high && arr[mid + 1] < arr[mid])
    {
        return arr[mid];
    }
  
    // Check if element at (mid - 1) 
    // is maximum element
    // Consider the cases like {4, 5, 1, 2, 3}
    if (mid > low && arr[mid] < arr[mid - 1])
    {
        return arr[mid - 1];
    }
  
    // Decide whether we need to go to
    // the left half or the right half
    if (arr[low] > arr[mid])
    {
        return findMax(arr, low, mid - 1);
    }
    else
    {
        return findMax(arr, mid + 1, high);
    }
}
  
// Driver code
public static void Main()
{
    int []arr = { 5, 6, 1, 2, 3, 4 };
    int n = arr.Length;
      
    Console.WriteLine(findMax(arr, 0, n - 1));
}
}
  
// This code is contributed by Ryuga

PHP

 $low && $arr[$mid] < $arr[$mid - 1])
    {
        return $arr[$mid - 1];
    }
  
    // Decide whether we need to go to
    // the left half or the right half
    if ($arr[$low] > $arr[$mid])
    {
        return findMax($arr, $low, $mid - 1);
    }
    else
    {
        return findMax($arr, $mid + 1, $high);
    }
}
  
// Driver code
$arr = array(5, 6, 1, 2, 3, 4);
$n = sizeof($arr);
echo findMax($arr, 0, $n - 1);
  
// This code is contributed
// by Akanksha Rai

输出：

6

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