本文概述

• C ++
• C
• Java
• Python3
• C#
• PHP
• C ++
• C
• Java
• Python3
• C#
• PHP

给定一个整数数组, 该数组最初是递增的, 然后递减的, 请在该数组中找到最大值。
例子 ：

Input: arr[] = {8, 10, 20, 80, 100, 200, 400, 500, 3, 2, 1} Output: 500Input: arr[] = {1, 3, 50, 10, 9, 7, 6} Output: 50Corner case (No decreasing part) Input: arr[] = {10, 20, 30, 40, 50} Output: 50Corner case (No increasing part) Input: arr[] = {120, 100, 80, 20, 0} Output: 120

推荐：请在"实践首先, 在继续解决方案之前。
方法1(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 < bits/stdc++.h> 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 < stdio.h> // 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 // elementclass 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 elementdef 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

的PHP

< ?php // PHP program to Find the maximum // element in an array which is first // increasing and then decreasingfunction findMaximum( $arr , $low , $high ) { $max = $arr [ $low ]; $i ; for ( $i = $low ; $i < = $high ; $i ++) { if ( $arr [ $i ] > $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. ?>

输出：

The maximum element is 70

时间复杂度：上)
【算法（在先增加然后减少的数组中找到最大元素）】方法2(二进制搜索)
我们可以为给定类型的数组修改标准的二进制搜索算法。
i)如果mid元素大于其两个相邻元素, 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。数组示例：{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。数组示例：{2, 4, 6, 8, 8, 10, 3, 1}
C ++

#include < bits/stdc++.h> 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 < stdio.h> 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 // elementclass 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 )); } }

Python3

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

的PHP

< ?php // PHP program to Find the maximum // element in an array which is // first increasing and then decreasingfunction 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 ]; /*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. ?>

输出：

The maximum element is 50

时间复杂度：O(Log n)
此方法仅适用于不同的数字。例如, 它不适用于{0, 1, 1, 2, 2, 2, 2, 2, 2, 4, 4, 4, 5, 3, 3, 2, 2, 1, 1}这样的数组。
如果发现任何不正确的地方, 或者想分享有关上述主题的更多信息, 请写评论。

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