给定数组arr[]，找到最大j – i，使得arr[j]大于arr[i]

2021年4月7日18:49:25 发表评论 1,325 次浏览

本文概述

给定一个数组arr[], 找到最大j – i, 使得arr[j] > arr[i]。

例子 ：

Input: {34, 8, 10, 3, 2, 80, 30, 33, 1}
  Output: 6  (j = 7, i = 1)

  Input: {9, 2, 3, 4, 5, 6, 7, 8, 18, 0}
  Output: 8 ( j = 8, i = 0)

  Input:  {1, 2, 3, 4, 5, 6}
  Output: 5  (j = 5, i = 0)

  Input:  {6, 5, 4, 3, 2, 1}
  Output: -1

方法1(简单但效率低下)

运行两个循环。在外部循环中, 从左开始一个接一个地选择元素。在内部循环中, 将拾取的元素与从右侧开始的元素进行比较。当你看到一个大于选取的元素的元素时, 请停止内部循环, 并保持当前的最大j-i更新。

C++

#include <bits/stdc++.h>
 
using namespace std;
/* For a given array arr[], returns the maximum j – i such that
     arr[j]> arr[i] */
int maxIndexDiff( int arr[], int n)
{
     int maxDiff = -1;
     int i, j;
 
     for (i = 0; i <n; ++i) {
         for (j = n - 1; j> i; --j) {
             if (arr[j]> arr[i] && maxDiff <(j - i))
                 maxDiff = j - i;
         }
     }
 
     return maxDiff;
}
 
int main()
{
     int arr[] = { 9, 2, 3, 4, 5, 6, 7, 8, 18, 0 };
     int n = sizeof (arr) /sizeof (arr[0]);
     int maxDiff = maxIndexDiff(arr, n);
     cout <<"\n"
          <<maxDiff;
     return 0;
}
 
//This code is contributed
//by Akanksha Rai(Abby_akku)

C

#include <stdio.h>
/* For a given array arr[], returns the maximum j – i such that
     arr[j]> arr[i] */
int maxIndexDiff( int arr[], int n)
{
     int maxDiff = -1;
     int i, j;
 
     for (i = 0; i <n; ++i) {
         for (j = n - 1; j> i; --j) {
             if (arr[j]> arr[i] && maxDiff <(j - i))
                 maxDiff = j - i;
         }
     }
 
     return maxDiff;
}
 
int main()
{
     int arr[] = { 9, 2, 3, 4, 5, 6, 7, 8, 18, 0 };
     int n = sizeof (arr) /sizeof (arr[0]);
     int maxDiff = maxIndexDiff(arr, n);
     printf ( "\n %d" , maxDiff);
     getchar ();
     return 0;
}

Java

class FindMaximum {
     /* For a given array arr[], returns the maximum j-i such that
        arr[j]> arr[i] */
     int maxIndexDiff( int arr[], int n)
     {
         int maxDiff = - 1 ;
         int i, j;
 
         for (i = 0 ; i <n; ++i) {
             for (j = n - 1 ; j> i; --j) {
                 if (arr[j]> arr[i] && maxDiff <(j - i))
                     maxDiff = j - i;
             }
         }
 
         return maxDiff;
     }
 
     /* Driver program to test above functions */
     public static void main(String[] args)
     {
         FindMaximum max = new FindMaximum();
         int arr[] = { 9 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 18 , 0 };
         int n = arr.length;
         int maxDiff = max.maxIndexDiff(arr, n);
         System.out.println(maxDiff);
     }
}

Python3

# Python3 program to find the maximum
# j – i such that arr[j]> arr[i]
  
# For a given array arr[], returns
# the maximum j – i such that
# arr[j]> arr[i]
def maxIndexDiff(arr, n):
     maxDiff = - 1
     for i in range ( 0 , n):
         j = n - 1
         while (j> i):
             if arr[j]> arr[i] and maxDiff <(j - i):
                 maxDiff = j - i;
             j - = 1
     
     return maxDiff
 
# driver code
arr = [ 9 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 18 , 0 ]
n = len (arr)
maxDiff = maxIndexDiff(arr, n)
print (maxDiff)
 
# This article is contributed by Smitha Dinesh Semwal

C#

//C# program to find the maximum
//j – i such that arr[j]> arr[i]
using System;
class GFG {
     //For a given array arr[], returns
     //the maximum j-i such that arr[j]> arr[i]
     static int maxIndexDiff( int [] arr, int n)
     {
         int maxDiff = -1;
         int i, j;
 
         for (i = 0; i <n; ++i) {
             for (j = n - 1; j> i; --j) {
                 if (arr[j]> arr[i] && maxDiff <(j - i))
                     maxDiff = j - i;
             }
         }
 
         return maxDiff;
     }
 
     //Driver program
     public static void Main()
     {
 
         int [] arr = { 9, 2, 3, 4, 5, 6, 7, 8, 18, 0 };
         int n = arr.Length;
         int maxDiff = maxIndexDiff(arr, n);
         Console.Write(maxDiff);
     }
}
//This Code is Contributed by Sam007

PHP

<?php
//PHP program to find the maximum
//j – i such that arr[j]> arr[i]
 
//For a given array arr[], returns
//the maximum j – i such that
//arr[j]> arr[i]
function maxIndexDiff( $arr , $n )
{
     $maxDiff = -1;
     
     for ( $i = 0; $i <$n ; ++ $i )
     {
         for ( $j = $n - 1; $j> $i ; -- $j )
         {
             if ( $arr[ $j ]> $arr[ $i ] &&
                $maxDiff <( $j - $i ))
                 $maxDiff = $j - $i ;
         }
     }
 
     return $maxDiff ;
}
 
//Driver Code
$arr = array (9, 2, 3, 4, 5, 6, 7, 8, 18, 0);
$n = count ( $arr );
$maxDiff = maxIndexDiff( $arr , $n );
echo $maxDiff ;
 
//This code is contributed by Sam007
?>

输出如下：

8

时间复杂度：O(n ^ 2)

方法2 –

改进蛮力算法并查找BUD, 即瓶颈, 不必要和重复的作品。快速观察实际上表明, 我们一直在寻找从数组末尾到当前索引的第一个最大元素。我们可以看到, 我们试图一次又一次地为数组中的每个元素找到最大的元素。假设我们有一个数组, 例如[1, 5, 12, 4, 9], 现在我们知道9是大于1、5和4的元素, 但是为什么我们需要一次又一次地找到它。实际上, 我们可以跟踪从数组结尾到开头的最大数量。这种方法将帮助我们更好地理解, 并且即兴采访也很容易。

方法：

  1. 从末尾遍历数组, 并跟踪当前索引右边的最大数字, 包括self
  2. 现在我们有了一个单调的递减数组, 并且我们知道可以使用二进制搜索找到最右边的较大元素的索引
  3. 现在, 我们将仅对数组中的每个元素使用二进制搜索, 并存储索引的最大差值, 就是这样。

C++

/* For a given array arr[], calculates the maximum j – i such that
     arr[j]> arr[i] */
#include <bits/stdc++.h>
using namespace std;
 
int main()
{
     vector<long long int> v{ 34, 8, 10, 3, 2, 80, 30, 33, 1 };
     int n = v.size();
     vector<long long int> maxFromEnd(n + 1, INT_MIN);
 
     //create an array maxfromEnd
     for ( int i = v.size() - 1; i>= 0; i--) {
         maxFromEnd[i] = max(maxFromEnd[i + 1], v[i]);
     }
 
     int result = 0;
 
     for ( int i = 0; i <v.size(); i++) {
         int low = i + 1, high = v.size() - 1, ans = i;
 
         while (low <= high) {
             int mid = (low + high) /2;
 
             if (v[i] <= maxFromEnd[mid]) {
                 //we store this as current answer and look
                 //for further larger number to the right side
                 ans = max(ans, mid);
                 low = mid + 1;
             }
             else {
                 high = mid - 1;
             }
         }
         //keeping a track of the
         //maximum difference in indices
         result = max(result, ans - i);
     }
     cout <<result <<endl;
}

Java

//Java program to implement
//the above approach
 
//For a given array arr[], //calculates the maximum j – i
//such that arr[j]> arr[i]
import java.util.*;
class GFG{
 
public static void main(String[] args)
{
   int []v = { 34 , 8 , 10 , 3 , 2 , 80 , 30 , 33 , 1 };
   int n = v.length;
   int []maxFromEnd = new int [n + 1 ];
   Arrays.fill(maxFromEnd, Integer.MIN_VALUE);
 
   //Create an array maxfromEnd
   for ( int i = v.length - 1 ; i>= 0 ; i--)
   {
     maxFromEnd[i] = Math.max(maxFromEnd[i + 1 ], v[i]);
   }
 
   int result = 0 ;
 
   for ( int i = 0 ; i <v.length; i++)
   {
     int low = i + 1 , high = v.length - 1 , ans = i;
 
     while (low <= high)
     {
       int mid = (low + high) /2 ;
 
       if (v[i] <= maxFromEnd[mid])
       {
         //We store this as current
         //answer and look for further
         //larger number to the right side
         ans = Math.max(ans, mid);
         low = mid + 1 ;
       }
       else
       {
         high = mid - 1 ;
       }
     }
     
     //Keeping a track of the
     //maximum difference in indices
     result = Math.max(result, ans - i);
   }
   System.out.print(result + "\n" );
}
}
 
//This code is contributed by shikhasingrajput

Python3

# Python3 program to implement
# the above approach
 
# For a given array arr, # calculates the maximum j – i
# such that arr[j]> arr[i]
 
# Driver code
if __name__ = = '__main__' :
   
     v = [ 34 , 8 , 10 , 3 , 2 , 80 , 30 , 33 , 1 ];
     n = len (v);
     maxFromEnd = [ - 38749432 ] * (n + 1 );
 
     # Create an array maxfromEnd
     for i in range (n - 1 , 0 , - 1 ):
         maxFromEnd[i] = max (maxFromEnd[i + 1 ], v[i]);
 
     result = 0 ;
 
     for i in range ( 0 , n):
         low = i + 1 ; high = n - 1 ; ans = i;
 
         while (low <= high):
             mid = int ((low + high) /2 );
 
             if (v[i] <= maxFromEnd[mid]):
               
                 # We store this as current
                 # answer and look for further
                 # larger number to the right side
                 ans = max (ans, mid);
                 low = mid + 1 ;
             else :
                 high = mid - 1 ;       
 
         # Keeping a track of the
         # maximum difference in indices
         result = max (result, ans - i);
     
     print (result, end = "");
     
# This code is contributed by Rajput-Ji

C#

//C# program to implement
//the above approach
 
//For a given array []arr, //calculates the maximum j – i
//such that arr[j]> arr[i]
using System;
class GFG{
 
public static void Main(String[] args)
{
   int []v = {34, 8, 10, 3, 2, 80, 30, 33, 1};
   int n = v.Length;
   int []maxFromEnd = new int [n + 1];
   
   for ( int i = 0;
            i <maxFromEnd.Length; i++)
     maxFromEnd[i] = int .MinValue;
 
   //Create an array maxfromEnd
   for ( int i = v.Length - 1;
            i>= 0; i--)
   {
     maxFromEnd[i] = Math.Max(maxFromEnd[i + 1], v[i]);
   }
 
   int result = 0;
 
   for ( int i = 0; i <v.Length; i++)
   {
     int low = i + 1, high = v.Length - 1, ans = i;
 
     while (low <= high)
     {
       int mid = (low + high) /2;
 
       if (v[i] <= maxFromEnd[mid])
       {
         //We store this as current
         //answer and look for further
         //larger number to the right side
         ans = Math.Max(ans, mid);
         low = mid + 1;
       }
       else
       {
         high = mid - 1;
       }
     }
 
     //Keeping a track of the
     //maximum difference in indices
     result = Math.Max(result, ans - i);
   }
   Console.Write(result + "\n" );
}
}
 
//This code is contributed by shikhasingrajput

输出如下：

6

时间复杂度：

O(N * log(N))

空间复杂度：O(n)

方法3 O(nLgn)

在特别注意重复项之后, 使用哈希和排序以小于二次复杂度的方式解决此问题。

方法：

  1. 遍历数组并将每个元素的索引存储在列表中(以处理重复项)。
  2. 对数组进行排序。
  3. 现在遍历数组, 并跟踪i和j的最大差。
  4. 对于j, 请考虑该元素可能的索引列表中的最后一个索引, 而对于我, 请考虑该列表中的第一个索引。 (因为索引是按升序附加的)。
  5. 不断更新最大差值, 直到数组末尾。

Python3

# Python3 implementation of the above approach
n = 9
a = [ 34 , 8 , 10 , 3 , 2 , 80 , 30 , 33 , 1 ]
 
# To store the index of an element.
index = dict ()
for i in range (n):
     if a[i] in index:
 
         # append to list (for duplicates)
         index[a[i]].append(i) 
     else :
 
         # if first occurrence
         index[a[i]] = [i]  
 
# sort the input array
a.sort()    
maxDiff = 0
 
# Temporary variable to keep track of minimum i
temp = n    
for i in range (n):
     if temp> index[a[i]][ 0 ]:
         temp = index[a[i]][ 0 ]
     maxDiff = max (maxDiff, index[a[i]][ - 1 ] - temp)
 
print (maxDiff)

输出如下：

6

时间复杂度：O(N * log(N))

方法4(有效)

为了解决这个问题, 我们需要获得arr[]的两个最佳索引：左索引i和右索引j。对于元素arr[i], 如果arr[i]左侧的元素小于arr[i], 则无需为左索引考虑arr[i]。同样, 如果arr[j]的右侧有一个更大的元素, 那么对于正确的索引, 我们不需要考虑这个j。因此, 我们构造了两个辅助数组LMin []和RMax [], 使得LMin [i]在包含arr[i]的arr[i]左侧保留最小的元素, 而RMax [j]在arr[i]右侧保留最大的元素arr[j]包括arr[j]。构建完这两个辅助数组后, 我们从左到右遍历这两个数组。在遍历LMin []和RMa []时, 如果我们看到LMin [i]大于RMax [j], 则必须向前移动LMin [](或执行i ++), 因为LMin [i]左侧的所有元素都是大于或等于LMin [i]。否则, 我们必须继续前进RMax [j], 以寻找更大的j – i值。

感谢celicom为这种方法建议算法。

C++

#include <bits/stdc++.h>
using namespace std;
 
/* For a given array arr[], returns the maximum j – i such that
    arr[j]> arr[i] */
int maxIndexDiff( int arr[], int n)
{
     int maxDiff;
     int i, j;
 
     int * LMin = new int [( sizeof ( int ) * n)];
     int * RMax = new int [( sizeof ( int ) * n)];
 
     /* Construct LMin[] such that
     LMin[i] stores the minimum value
     from (arr[0], arr[1], ... arr[i]) */
     LMin[0] = arr[0];
     for (i = 1; i <n; ++i)
         LMin[i] = min(arr[i], LMin[i - 1]);
 
     /* Construct RMax[] such that
     RMax[j] stores the maximum value from
     (arr[j], arr[j+1], ..arr[n-1]) */
     RMax[n - 1] = arr[n - 1];
     for (j = n - 2; j>= 0; --j)
         RMax[j] = max(arr[j], RMax[j + 1]);
 
     /* Traverse both arrays from left to right
     to find optimum j - i. This process is similar to
     merge() of MergeSort */
     i = 0, j = 0, maxDiff = -1;
     while (j <n && i <n) {
         if (LMin[i] <RMax[j]) {
             maxDiff = max(maxDiff, j - i);
             j = j + 1;
         }
         else
             i = i + 1;
     }
 
     return maxDiff;
}
 
//Driver Code
int main()
{
     int arr[] = { 9, 2, 3, 4, 5, 6, 7, 8, 18, 0 };
     int n = sizeof (arr) /sizeof (arr[0]);
     int maxDiff = maxIndexDiff(arr, n);
     cout <<maxDiff;
     return 0;
}
 
//This code is contributed by rathbhupendra

C

#include <stdio.h>
 
/* Utility Functions to get max and minimum of two integers */
int max( int x, int y)
{
     return x> y ? x : y;
}
 
int min( int x, int y)
{
     return x <y ? x : y;
}
 
/* For a given array arr[], returns the maximum j – i such that
     arr[j]> arr[i] */
int maxIndexDiff( int arr[], int n)
{
     int maxDiff;
     int i, j;
 
     int * LMin = ( int *) malloc ( sizeof ( int ) * n);
     int * RMax = ( int *) malloc ( sizeof ( int ) * n);
 
     /* Construct LMin[] such that LMin[i] stores the minimum value
        from (arr[0], arr[1], ... arr[i]) */
     LMin[0] = arr[0];
     for (i = 1; i <n; ++i)
         LMin[i] = min(arr[i], LMin[i - 1]);
 
     /* Construct RMax[] such that RMax[j] stores the maximum value
        from (arr[j], arr[j+1], ..arr[n-1]) */
     RMax[n - 1] = arr[n - 1];
     for (j = n - 2; j>= 0; --j)
         RMax[j] = max(arr[j], RMax[j + 1]);
 
     /* Traverse both arrays from left to right to find optimum j - i
         This process is similar to merge() of MergeSort */
     i = 0, j = 0, maxDiff = -1;
     while (j <n && i <n) {
         if (LMin[i] <RMax[j]) {
             maxDiff = max(maxDiff, j - i);
             j = j + 1;
         }
         else
             i = i + 1;
     }
 
     return maxDiff;
}
 
/* Driver program to test above functions */
int main()
{
     int arr[] = { 9, 2, 3, 4, 5, 6, 7, 8, 18, 0 };
     int n = sizeof (arr) /sizeof (arr[0]);
     int maxDiff = maxIndexDiff(arr, n);
     printf ( "\n %d" , maxDiff);
     getchar ();
     return 0;
}

Java

class FindMaximum {
     /* Utility Functions to get max and minimum of two integers */
     int max( int x, int y)
     {
         return x> y ? x : y;
     }
 
     int min( int x, int y)
     {
         return x <y ? x : y;
     }
 
     /* For a given array arr[], returns the maximum j-i such that
        arr[j]> arr[i] */
     int maxIndexDiff( int arr[], int n)
     {
         int maxDiff;
         int i, j;
 
         int RMax[] = new int [n];
         int LMin[] = new int [n];
 
         /* Construct LMin[] such that LMin[i] stores the minimum value
            from (arr[0], arr[1], ... arr[i]) */
         LMin[ 0 ] = arr[ 0 ];
         for (i = 1 ; i <n; ++i)
             LMin[i] = min(arr[i], LMin[i - 1 ]);
 
         /* Construct RMax[] such that RMax[j] stores the maximum value
            from (arr[j], arr[j+1], ..arr[n-1]) */
         RMax[n - 1 ] = arr[n - 1 ];
         for (j = n - 2 ; j>= 0 ; --j)
             RMax[j] = max(arr[j], RMax[j + 1 ]);
 
         /* Traverse both arrays from left to right to find optimum j - i
            This process is similar to merge() of MergeSort */
         i = 0 ;
         j = 0 ;
         maxDiff = - 1 ;
         while (j <n && i <n) {
             if (LMin[i] <RMax[j]) {
                 maxDiff = max(maxDiff, j - i);
                 j = j + 1 ;
             }
             else
                 i = i + 1 ;
         }
 
         return maxDiff;
     }
 
     /* Driver program to test the above functions */
     public static void main(String[] args)
     {
         FindMaximum max = new FindMaximum();
         int arr[] = { 9 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 18 , 0 };
         int n = arr.length;
         int maxDiff = max.maxIndexDiff(arr, n);
         System.out.println(maxDiff);
     }
}

Python3

# Utility Functions to get max
# and minimum of two integers
def max (a, b):
     if (a> b):
         return a
     else :
         return b
 
def min (a, b):
     if (a <b):
         return a
     else :
         return b
 
# For a given array arr[], # returns the maximum j - i
# such that arr[j]> arr[i]
def maxIndexDiff(arr, n):
     maxDiff = 0 ;
     LMin = [ 0 ] * n
     RMax = [ 0 ] * n
 
     # Construct LMin[] such that
     # LMin[i] stores the minimum
     # value from (arr[0], arr[1], # ... arr[i])
     LMin[ 0 ] = arr[ 0 ]
     for i in range ( 1 , n):
         LMin[i] = min (arr[i], LMin[i - 1 ])
 
     # Construct RMax[] such that
     # RMax[j] stores the maximum
     # value from (arr[j], arr[j + 1], # ..arr[n-1])
     RMax[n - 1 ] = arr[n - 1 ]
     for j in range (n - 2 , - 1 , - 1 ):
         RMax[j] = max (arr[j], RMax[j + 1 ]);
 
     # Traverse both arrays from left
     # to right to find optimum j - i
     # This process is similar to
     # merge() of MergeSort
     i, j = 0 , 0
     maxDiff = - 1
     while (j <n and i <n):
         if (LMin[i] <RMax[j]):
             maxDiff = max (maxDiff, j - i)
             j = j + 1
         else :
             i = i + 1
 
     return maxDiff
 
# Driver Code
if (__name__ = = '__main__' ):
     arr = [ 9 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 18 , 0 ]
     n = len (arr)
     maxDiff = maxIndexDiff(arr, n)
     print (maxDiff)
 
# This code is contributed
# by gautam karakoti

C#

//C# program to find the maximum
//j – i such that arr[j]> arr[i]
using System;
 
class GFG {
     //Utility Functions to get max
     //and minimum of two integers
     static int max( int x, int y)
     {
         return x> y ? x : y;
     }
 
     static int min( int x, int y)
     {
         return x <y ? x : y;
     }
 
     //For a given array arr[], returns
     //the maximum j-i such thatarr[j]> arr[i]
     static int maxIndexDiff( int [] arr, int n)
     {
         int maxDiff;
         int i, j;
 
         int [] RMax = new int [n];
         int [] LMin = new int [n];
 
         //Construct LMin[] such that LMin[i]
         //stores the minimum value
         //from (arr[0], arr[1], ... arr[i])
         LMin[0] = arr[0];
         for (i = 1; i <n; ++i)
             LMin[i] = min(arr[i], LMin[i - 1]);
 
         //Construct RMax[] such that
         //RMax[j] stores the maximum value
         //from (arr[j], arr[j+1], ..arr[n-1])
         RMax[n - 1] = arr[n - 1];
         for (j = n - 2; j>= 0; --j)
             RMax[j] = max(arr[j], RMax[j + 1]);
 
         //Traverse both arrays from left
         //to right to find optimum j - i
         //This process is similar to merge()
         //of MergeSort
         i = 0;
         j = 0;
         maxDiff = -1;
         while (j <n && i <n) {
             if (LMin[i] <RMax[j]) {
                 maxDiff = max(maxDiff, j - i);
                 j = j + 1;
             }
             else
                 i = i + 1;
         }
 
         return maxDiff;
     }
 
     //Driver program
     public static void Main()
     {
 
         int [] arr = { 9, 2, 3, 4, 5, 6, 7, 8, 18, 0 };
         int n = arr.Length;
         int maxDiff = maxIndexDiff(arr, n);
         Console.Write(maxDiff);
     }
}
//This Code is Contributed by Sam007

PHP

<?php
//PHP program to find the maximum
//j – i such that arr[j]> arr[i]
 
//For a given array arr[], //returns the maximum j - i
//such that arr[j]> arr[i]
function maxIndexDiff( $arr , $n )
{
     $maxDiff = 0;
     $LMin = array_fill (0, $n , NULL);
     $RMax = array_fill (0, $n , NULL);
 
     //Construct LMin[] such that
     //LMin[i] stores the minimum
     //value from (arr[0], arr[1], //... arr[i])
     $LMin [0] = $arr[0];
     for ( $i = 1; $i <$n ; $i ++)
         $LMin [ $i ] = min( $arr[ $i ], $LMin [ $i - 1]);
 
     //Construct RMax[] such that
     //RMax[j] stores the maximum
     //value from (arr[j], arr[j+1], //..arr[n-1])
     $RMax [ $n - 1] = $arr[ $n - 1];
     for ( $j = $n - 2; $j>= 0; $j --)
         $RMax [ $j ] = max( $arr[ $j ], $RMax [ $j + 1]);
 
     //Traverse both arrays from left
     //to right to find optimum j - i
     //This process is similar to
     //merge() of MergeSort
     $i = 0;
     $j = 0;
     $maxDiff = -1;
     while ( $j <$n && $i <$n )
         if ( $LMin [ $i ] <$RMax [ $j ])
         {
             $maxDiff = max( $maxDiff , $j - $i );
             $j = $j + 1;
         }
         else
             $i = $i + 1;
 
     return $maxDiff ;
}
 
//Driver Code
$arr = array (9, 2, 3, 4, 5, 6, 7, 8, 18, 0);
$n = sizeof( $arr );
$maxDiff = maxIndexDiff( $arr , $n );
echo $maxDiff ;
 
//This code is contributed
//by ChitraNayal
?>

输出如下：

8

时间复杂度：O(n)

辅助空间：O(n)

如果你发现上述代码/算法有误, 请写评论, 或者找到其他解决相同问题的方法。

木子山

发表评论

:?: :razz: :sad: :evil: :!: :smile: :oops: :grin: :eek: :shock: :???: :cool: :lol: :mad: :twisted: :roll: :wink: :idea: :arrow: :neutral: :cry: :mrgreen: