找出数组中先增加后减少的最大元素

给定一个最初增加然后减少的整数数组，找到数组中的最大值。
例子 ：

Input: arr[] = {8, 10, 20, 80, 100, 200, 400, 500, 3, 2, 1}
Output: 500

Input: arr[] = {1, 3, 50, 10, 9, 7, 6}
Output: 50

Corner case (No decreasing part)
Input: arr[] = {10, 20, 30, 40, 50}
Output: 50

Corner case (No increasing part)
Input: arr[] = {120, 100, 80, 20, 0}
Output: 120

方法一(线性搜索)
我们可以遍历数组并跟踪最大值和元素。最后返回最大元素。

C++

// C++ program to find maximum
// element
#include 
using namespace std;
 
// function to find the maximum element
int findMaximum(int arr[], int low, int high)
{
    int max = arr[low];
    int i;
    for (i = low + 1; i <= high; i++)
    {
        if (arr[i] > max)
            max = arr[i];
         
        // break when once an element is smaller than
        // the max then it will go on decreasing
        // and no need to check after that
        else
            break;
    }
    return max;
}
 
/* Driver code*/
int main()
{
    int arr[] = {1, 30, 40, 50, 60, 70, 23, 20};
    int n = sizeof(arr)/sizeof(arr[0]);
    cout << "The maximum element is " << findMaximum(arr, 0, n-1);
    return 0;
}
 
// This is code is contributed by rathbhupendra

C

// C program to find maximum
// element
#include 
 
// function to find the maximum element
int findMaximum(int arr[], int low, int high)
{
int max = arr[low];
int i;
for (i = low+1; i <= high; i++)
{
    if (arr[i] > max)
        max = arr[i];
// break when once an element is smaller than
// the max then it will go on decreasing
// and no need to check after that
    else
        break;
}
return max;
}
 
/* Driver program to check above functions */
int main()
{
int arr[] = {1, 30, 40, 50, 60, 70, 23, 20};
int n = sizeof(arr)/sizeof(arr[0]);
printf("The maximum element is %d", findMaximum(arr, 0, n-1));
getchar();
return 0;
}

Java

// java program to find maximum
// element
 
class Main
{  
    // function to find the
    // maximum element
    static int findMaximum(int arr[], int low, int high)
    {
       int max = arr[low];
       int i;
       for (i = low; i <= high; i++)
       {
           if (arr[i] > max)
              max = arr[i];
       }
       return max;
    }
     
    // main function
    public static void main (String[] args)
    {
        int arr[] = {1, 30, 40, 50, 60, 70, 23, 20};
        int n = arr.length;
        System.out.println("The maximum element is "+
                            findMaximum(arr, 0, n-1));
    }
}

Python3

# Python3 program to find
# maximum element
 
def findMaximum(arr, low, high):
    max = arr[low]
    i = low
    for i in range(high+1):
        if arr[i] > max:
            max = arr[i]
    return max
 
# Driver program to check above functions */
arr = [1, 30, 40, 50, 60, 70, 23, 20]
n = len(arr)
print ("The maximum element is %d"%
        findMaximum(arr, 0, n-1))
 
# This code is contributed by Shreyanshi Arun.

C#

// C# program to find maximum
// element
using System;
 
class GFG
{
    // function to find the
    // maximum element
    static int findMaximum(int []arr, int low, int high)
    {
        int max = arr[low];
        int i;
        for (i = low; i <= high; i++)
        {
            if (arr[i] > max)
                max = arr[i];
        }
        return max;
    }
     
    // Driver code
    public static void Main ()
    {
        int []arr = {1, 30, 40, 50, 60, 70, 23, 20};
        int n = arr.Length;
        Console.Write("The maximum element is "+
                        findMaximum(arr, 0, n-1));
    }
}
 
// This code is contributed by Sam007

PHP

 $max)
        $max = $arr[$i];
}
return $max;
}
 
// Driver Code
$arr = array(1, 30, 40, 50,
             60, 70, 23, 20);
$n = count($arr);
echo "The maximum element is ",
      findMaximum($arr, 0, $n-1);
 
// This code is contributed by anuj_67.
?>

Javascript

 
// Javascript program to find maximum
// element
 
// function to find the maximum element
function findMaximum(arr, low, high)
{
    var max = arr[low];
    var i;
    for (i = low + 1; i <= high; i++)
    {
        if (arr[i] > max)
            max = arr[i];
         
        // break when once an element is smaller than
        // the max then it will go on decreasing
        // and no need to check after that
        else
            break;
    }
    return max;
}
 
/* Driver code*/
var arr = [1, 30, 40, 50, 60, 70, 23, 20];
var n = arr.length;
document.write("The maximum element is " + findMaximum(arr, 0, n-1));
 

输出

The maximum element is 70

时间复杂度： O(n)
方法二(二分查找)
我们可以修改给定类型数组的标准二分搜索算法。
i) 如果 mid 元素大于它的两个相邻元素，则 mid 是最大值。
ii) 如果 mid 元素大于其下一个元素且小于前一个元素，则最大值位于 mid 的左侧。示例数组：{3, 50, 10, 9, 7, 6}
iii) 如果 mid 元素小于其下一个元素且大于前一个元素，则最大值位于 mid 的右侧。示例数组：{2, 4, 6, 8, 10, 3, 1}

C++

#include 
using namespace std;
 
int findMaximum(int arr[], int low, int high)
{
 
    /* Base Case: Only one element is present in arr[low..high]*/
    if (low == high)
        return arr[low];
     
    /* If there are two elements and first is greater then
        the first element is maximum */
    if ((high == low + 1) && arr[low] >= arr[high])
        return arr[low];
     
    /* If there are two elements and second is greater then
        the second element is maximum */
    if ((high == low + 1) && arr[low] < arr[high])
        return arr[high];
     
    int mid = (low + high)/2; /*low + (high - low)/2;*/
     
    /* If we reach a point where arr[mid] is greater than both of
        its adjacent elements arr[mid-1] and arr[mid+1], then arr[mid]
        is the maximum element*/
    if ( arr[mid] > arr[mid + 1] && arr[mid] > arr[mid - 1])
        return arr[mid];
     
    /* If arr[mid] is greater than the next
        element and smaller than the previous
        element then maximum lies on left side of mid */
    if (arr[mid] > arr[mid + 1] && arr[mid] < arr[mid - 1])
        return findMaximum(arr, low, mid-1);
         
        // when arr[mid] is greater than arr[mid-1]
        // and smaller than arr[mid+1]
    else
        return findMaximum(arr, mid + 1, high);
}
 
/* Driver code */
int main()
{
    int arr[] = {1, 3, 50, 10, 9, 7, 6};
    int n = sizeof(arr)/sizeof(arr[0]);
    cout << "The maximum element is " << findMaximum(arr, 0, n-1);
    return 0;
}
 
// This is code is contributed by rathbhupendra

C

#include 
 
int findMaximum(int arr[], int low, int high)
{
 
   /* Base Case: Only one element is present in arr[low..high]*/
   if (low == high)
     return arr[low];
 
   /* If there are two elements and first is greater then
      the first element is maximum */
   if ((high == low + 1) && arr[low] >= arr[high])
      return arr[low];
 
   /* If there are two elements and second is greater then
      the second element is maximum */
   if ((high == low + 1) && arr[low] < arr[high])
      return arr[high];
 
   int mid = (low + high)/2;   /*low + (high - low)/2;*/
 
   /* If we reach a point where arr[mid] is greater than both of
     its adjacent elements arr[mid-1] and arr[mid+1], then arr[mid]
     is the maximum element*/
   if ( arr[mid] > arr[mid + 1] && arr[mid] > arr[mid - 1])
      return arr[mid];
 
   /* If arr[mid] is greater than the next element and smaller than the previous
    element then maximum lies on left side of mid */
   if (arr[mid] > arr[mid + 1] && arr[mid] < arr[mid - 1])
     return findMaximum(arr, low, mid-1);
   else // when arr[mid] is greater than arr[mid-1] and smaller than arr[mid+1]
     return findMaximum(arr, mid + 1, high);
}
 
/* Driver program to check above functions */
int main()
{
   int arr[] = {1, 3, 50, 10, 9, 7, 6};
   int n = sizeof(arr)/sizeof(arr[0]);
   printf("The maximum element is %d", findMaximum(arr, 0, n-1));
   getchar();
   return 0;
}

Java

// java program to find maximum
// element
 
class Main
{  
    // function to find the
    // maximum element
    static int findMaximum(int arr[], int low, int high)
    {
      
       /* Base Case: Only one element is
          present in arr[low..high]*/
       if (low == high)
         return arr[low];
      
       /* If there are two elements and
          first is greater then the first
          element is maximum */
       if ((high == low + 1) && arr[low] >= arr[high])
          return arr[low];
      
       /* If there are two elements and
          second is greater then the second
          element is maximum */
       if ((high == low + 1) && arr[low] < arr[high])
          return arr[high];
         
       /*low + (high - low)/2;*/
       int mid = (low + high)/2;  
      
       /* If we reach a point where arr[mid]
          is greater than both of its adjacent
          elements arr[mid-1] and arr[mid+1],
          then arr[mid] is the maximum element*/
       if ( arr[mid] > arr[mid + 1] && arr[mid] > arr[mid - 1])
          return arr[mid];
      
       /* If arr[mid] is greater than the next
          element and smaller than the previous
          element then maximum lies on left side
          of mid */
       if (arr[mid] > arr[mid + 1] && arr[mid] < arr[mid - 1])
         return findMaximum(arr, low, mid-1);
       else
         return findMaximum(arr, mid + 1, high);
    }
     
    // main function
    public static void main (String[] args)
    {
        int arr[] = {1, 3, 50, 10, 9, 7, 6};
        int n = arr.length;
        System.out.println("The maximum element is "+
                            findMaximum(arr, 0, n-1));
    }
}

蟒蛇3

def findMaximum(arr, low, high):
    # Base Case: Only one element is present in arr[low..high]*/
    if low == high:
        return arr[low]
  
    # If there are two elements and first is greater then
    # the first element is maximum */
    if high == low + 1 and arr[low] >= arr[high]:
        return arr[low];
  
    # If there are two elements and second is greater then
    # the second element is maximum */
    if high == low + 1 and arr[low] < arr[high]:
        return arr[high]
  
    mid = (low + high)//2   #low + (high - low)/2;*/
  
    # If we reach a point where arr[mid] is greater than both of
    # its adjacent elements arr[mid-1] and arr[mid+1], then arr[mid]
    # is the maximum element*/
    if arr[mid] > arr[mid + 1] and arr[mid] > arr[mid - 1]:
        return arr[mid]
  
    # If arr[mid] is greater than the next element and smaller than the previous
    # element then maximum lies on left side of mid */
    if arr[mid] > arr[mid + 1] and arr[mid] < arr[mid - 1]:
        return findMaximum(arr, low, mid-1)
    else: # when arr[mid] is greater than arr[mid-1] and smaller than arr[mid+1]
        return findMaximum(arr, mid + 1, high)
  
# Driver program to check above functions */
arr = [1, 3, 50, 10, 9, 7, 6]
n = len(arr)
print ("The maximum element is %d"% findMaximum(arr, 0, n-1))
 
# This code is contributed by Shreyanshi Arun.

C＃

// C# program to find maximum
// element
using System;
 
class GFG
{
    // function to find the
    // maximum element
    static int findMaximum(int []arr, int low, int high)
    {
     
    /* Base Case: Only one element is
        present in arr[low..high]*/
    if (low == high)
        return arr[low];
     
    /* If there are two elements and
        first is greater then the first
        element is maximum */
    if ((high == low + 1) && arr[low] >= arr[high])
        return arr[low];
     
    /* If there are two elements and
        second is greater then the second
        element is maximum */
    if ((high == low + 1) && arr[low] < arr[high])
        return arr[high];
         
    /*low + (high - low)/2;*/
    int mid = (low + high)/2;
     
    /* If we reach a point where arr[mid]
        is greater than both of its adjacent
        elements arr[mid-1] and arr[mid+1],
        then arr[mid] is the maximum element*/
    if ( arr[mid] > arr[mid + 1] && arr[mid] > arr[mid - 1])
        return arr[mid];
     
    /* If arr[mid] is greater than the next
        element and smaller than the previous
        element then maximum lies on left side
        of mid */
    if (arr[mid] > arr[mid + 1] && arr[mid] < arr[mid - 1])
        return findMaximum(arr, low, mid-1);
    else
        return findMaximum(arr, mid + 1, high);
    }
     
    // main function
    public static void Main()
    {
        int []arr = {1, 3, 50, 10, 9, 7, 6};
        int n = arr.Length;
        Console.Write("The maximum element is "+
                            findMaximum(arr, 0, n-1));
    }
}
// This code is contributed by Sam007

PHP

= $arr[$high])
        return $arr[$low];
     
    /* If there are two elements
       and second is greater then
       the second element is maximum */
    if (($high == $low + 1) &&
         $arr[$low] < $arr[$high])
        return $arr[$high];
     
    /*low + (high - low)/2;*/
    $mid = ($low + $high) / 2;
     
    /* If we reach a point where
       arr[mid] is greater than
       both of its adjacent elements
       arr[mid-1] and arr[mid+1],
       then arr[mid] is the maximum
       element */
    if ( $arr[$mid] > $arr[$mid + 1] &&
         $arr[$mid] > $arr[$mid - 1])
        return $arr[$mid];
     
    /* If arr[mid] is greater than
       the next element and smaller
       than the previous element then
       maximum lies on left side of mid */
    if ($arr[$mid] > $arr[$mid + 1] &&
        $arr[$mid] < $arr[$mid - 1])
        return findMaximum($arr, $low, $mid - 1);
     
    // when arr[mid] is greater than
    // arr[mid-1] and smaller than
    // arr[mid+1]   
    else
        return findMaximum($arr,
                           $mid + 1, $high);
}
 
// Driver Code
$arr = array(1, 3, 50, 10, 9, 7, 6);
$n = sizeof($arr);
echo("The maximum element is ");
echo(findMaximum($arr, 0, $n-1));
 
// This code is contributed by nitin mittal.
?>

Javascript

   
function findMaximum( arr, low, high)
{ 
   
    /* Base Case: Only one element is present in arr[low..high]*/
    if (low == high)
        return arr[low];
       
    /* If there are two elements and first is greater then
        the first element is maximum */
    if ((high == low + 1) && arr[low] >= arr[high])
        return arr[low];
       
    /* If there are two elements and second is greater then
        the second element is maximum */
    if ((high == low + 1) && arr[low] < arr[high])
        return arr[high];
       
     mid = (low + high)/2;
     /*low + (high - low)/2;*/
       
    /* If we reach a point where arr[mid] is greater than both of
        its adjacent elements arr[mid-1] and arr[mid+1], then arr[mid]
        is the maximum element*/
    if ( arr[mid] > arr[mid + 1] && arr[mid] > arr[mid - 1])
        return arr[mid];
       
    /* If arr[mid] is greater than the next
        element and smaller than the previous
        element then maximum lies on left side of mid */
    if (arr[mid] > arr[mid + 1] && arr[mid] < arr[mid - 1])
        return findMaximum(arr, low, mid-1);
           
        // when arr[mid] is greater than arr[mid-1]
        // and smaller than arr[mid+1]
        return findMaximum(arr, mid + 1, high);
}
   
/* Driver code */
  
     arr = new Array(1, 3, 50, 10, 9, 7, 6);
     n = arr.length;
document.write("The maximum element is" + "\n" + findMaximum(arr, 0, n-1));
 
// This code is contributed by simranarora5sos
 

输出

The maximum element is 50

方法三(二分搜索——迭代解)

二分搜索的迭代方法，以找到数组中先增加然后减少的最大元素。

我们可以修改给定类型数组的标准二分搜索算法。

i) 如果 mid 元素大于它的两个相邻元素，则 mid 是最大值。

ii) 如果 mid 元素大于其下一个元素且小于前一个元素，则最大值位于 mid 的左侧。

示例数组：{3, 50, 10, 9, 7, 6}

iii) 如果 mid 元素小于其下一个元素且大于前一个元素，则最大值位于 mid 的右侧。示例数组：{2, 4, 6, 8, 10, 3, 1}

C++

#include 
using namespace std;
 
int maxInBitonic(int arr[], int l, int r)
{
 
    while (l <= r) {
 
        int m = l + (r - l) / 2; // m = (l + r) / 2
 
        /****Base Cases Starts*****/
 
        /* If there are two elements and first is greater
           then the first element is maximum */
        if ((r == l + 1) && arr[l] >= arr[r])
            return arr[l];
 
        /* If there are two elements and second is greater
           then the second element is maximum */
 
        if ((r == l + 1) && arr[l] < arr[r])
            return arr[r];
 
        /* If we reach a point where arr[mid] is greater
           than both of its adjacent elements arr[mid-1] and
           arr[mid+1], then arr[mid] is the maximum
           element*/
        if (arr[m] > arr[m + 1] && arr[m] > arr[m - 1])
            return arr[m];
 
        /****Base Case ends *****/
 
        // move to left with l and r=m-1
        if (arr[m] > arr[m + 1] && arr[m] < arr[m - 1])
            r = m - 1;
 
        else
            l = m + 1; // move to right with l=m+1 and r
    }
    // if we reach here, then element was
    // not present
    return -1;
}
 
// Driver function
int main()
{
    int arr[] = { 1, 3, 50, 10, 9, 7, 6 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << "The maximum element is "
         << maxInBitonic(arr, 0, n - 1);
    return 0;
}

Java

import java.util.*;
 
class GFG{
 
static int maxInBitonic(int arr[], int l, int r)
{
 
    while (l <= r) {
 
        int m = l + (r - l) / 2; // m = (l + r) / 2
 
        /****Base Cases Starts*****/
 
        /* If there are two elements and first is greater
           then the first element is maximum */
        if ((r == l + 1) && arr[l] >= arr[r])
            return arr[l];
 
        /* If there are two elements and second is greater
           then the second element is maximum */
 
        if ((r == l + 1) && arr[l] < arr[r])
            return arr[r];
 
        /* If we reach a point where arr[mid] is greater
           than both of its adjacent elements arr[mid-1] and
           arr[mid+1], then arr[mid] is the maximum
           element*/
        if (arr[m] > arr[m + 1] && arr[m] > arr[m - 1])
            return arr[m];
 
        /****Base Case ends *****/
 
        // move to left with l and r=m-1
        if (arr[m] > arr[m + 1] && arr[m] < arr[m - 1])
            r = m - 1;
 
        else
            l = m + 1; // move to right with l=m+1 and r
    }
    // if we reach here, then element was
    // not present
    return -1;
}
 
// Driver function
public static void main(String[] args)
{
    int arr[] = { 1, 3, 50, 10, 9, 7, 6 };
    int n = arr.length;
    System.out.print("The maximum element is "
         + maxInBitonic(arr, 0, n - 1));
}
}
 
// This code is contributed by todaysgaurav

C＃

using System;
 
class GFG{
 
static int maxInBitonic(int []arr, int l, int r)
{
 
    while (l <= r) {
 
        int m = l + (r - l) / 2; // m = (l + r) / 2
 
        /****Base Cases Starts*****/
 
        /* If there are two elements and first is greater
           then the first element is maximum */
        if ((r == l + 1) && arr[l] >= arr[r])
            return arr[l];
 
        /* If there are two elements and second is greater
           then the second element is maximum */
 
        if ((r == l + 1) && arr[l] < arr[r])
            return arr[r];
 
        /* If we reach a point where arr[mid] is greater
           than both of its adjacent elements arr[mid-1] and
           arr[mid+1], then arr[mid] is the maximum
           element*/
        if (arr[m] > arr[m + 1] && arr[m] > arr[m - 1])
            return arr[m];
 
        /****Base Case ends *****/
 
        // move to left with l and r=m-1
        if (arr[m] > arr[m + 1] && arr[m] < arr[m - 1])
            r = m - 1;
 
        else
            l = m + 1; // move to right with l=m+1 and r
    }
    // if we reach here, then element was
    // not present
    return -1;
}
 
// Driver function
public static void Main(String[] args)
{
    int []arr = { 1, 3, 50, 10, 9, 7, 6 };
    int n = arr.Length;
    Console.Write("The maximum element is "
         + maxInBitonic(arr, 0, n - 1));
}
}
 
// This code is contributed by shivanisinghss2110

Javascript

// JavaScript program
 
function maxInBitonic(arr, l, r)
{
 
    while (l <= r) {
 
        var m = l + (r - l) / 2; // m = (l + r) / 2
 
        /****Base Cases Starts*****/
 
        /* If there are two elements and first is greater
           then the first element is maximum */
        if ((r == l + 1) && arr[l] >= arr[r])
            return arr[l];
 
        /* If there are two elements and second is greater
           then the second element is maximum */
 
        if ((r == l + 1) && arr[l] < arr[r])
            return arr[r];
 
        /* If we reach a point where arr[mid] is greater
           than both of its adjacent elements arr[mid-1] and
           arr[mid+1], then arr[mid] is the maximum
           element*/
        if (arr[m] > arr[m + 1] && arr[m] > arr[m - 1])
            return arr[m];
 
        /****Base Case ends *****/
 
        // move to left with l and r=m-1
        if (arr[m] > arr[m + 1] && arr[m] < arr[m - 1])
            r = m - 1;
 
        else
            l = m + 1; // move to right with l=m+1 and r
    }
    // if we reach here, then element was
    // not present
    return -1;
}
 
// Driver function
    var arr = [ 1, 3, 50, 10, 9, 7, 6 ];
    var n = arr.length;
   document.write("The maximum element is "
         + maxInBitonic(arr, 0, n - 1));
 
 
// This code is contributed by shivanisinghss2110
 

输出

The maximum element is 50

时间复杂度： O(log n)
辅助空间： O(1)

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