给定数组中的最大定点（值等于索引）

给定一个大小为N的数组arr[] ，任务是找到最大索引i使得arr[i]等于i。如果数组arr[] 中没有这样的索引，则打印-1 。

例子：

Input: arr[ ] = {-10, -5, 0, 3, 7}
Output: 3
Explanation: Only for i=3, arr[3] = 3

Input: arr[ ] = {0, 2, 5, 8, 4}
Output: 4

方法：按照以下步骤解决此问题：

• 使用变量i在范围[N-1, 0] 中迭代：  
  • 如果当前元素等于i ，则打印i并返回。
• 如果没有这样的索引，则打印-1。

下面是上述方法的实现：

C++

// C++ implementation of the above approach
#include 
using namespace std;
  
// Function to  find the maximum index
// i such that arr[i] is equal to i
void findLargestIndex(int arr[], int n)
{
  
    // Traversing the array from
    // backwards
    for (int i = n - 1; i >= 0; i--) {
  
        // If arr[i] is equal to i
        if (arr[i] == i) {
            cout << i << endl;
            return;
        }
    }
  
    // If there is no such index
    cout << -1 << endl;
}
  
// Driver code
int main()
{
    // Given Input
    int arr[] = { -10, -5, 0, 3, 7 };
    int n = sizeof(arr) / sizeof(arr[0]);
  
    // Function Call
    findLargestIndex(arr, n);
    return 0;
}

Java

// Java implementation of the above approach
import java.io.*;
  
class GFG
{
    
    // Function to  find the maximum index
    // i such that arr[i] is equal to i
    static void findLargestIndex(int arr[], int n)
    {
  
        // Traversing the array from
        // backwards
        for (int i = n - 1; i >= 0; i--) {
  
            // If arr[i] is equal to i
            if (arr[i] == i) {
                System.out.println(i);
                return;
            }
        }
  
        // If there is no such index
        System.out.println(-1);
    }
  
    // Driver code
    public static void main(String[] args)
    {
        
        // Given Input
        int arr[] = { -10, -5, 0, 3, 7 };
        int n = arr.length;
  
        // Function Call
        findLargestIndex(arr, n);
    }
}
  
// This code is contributed by Potta Lokesh

Python

# Python implementation of the above approach
# Function to  find the maximum index
# i such that arr[i] is equal to i
def findLargestIndex(arr,  n):
  
    # Traversing the array from
    # backwards
    for i in range (n):
  
        # If arr[i] is equal to i
        if (arr[i] == i) :
            print( i )
            return
          
    # If there is no such index
    print( -1 )
  
# Driver code
# Given Input
arr =  [-10, -5, 0, 3, 7 ]
n = len(arr)
  
# Function Call
findLargestIndex(arr, n)
      
# This code is contributed by shivanisinghss2110

C#

// C# implementation of the above approach
using System;
  
class GFG
{
    
    // Function to  find the maximum index
    // i such that arr[i] is equal to i
    static void findLargestIndex(int []arr, int n)
    {
  
        // Traversing the array from
        // backwards
        for (int i = n - 1; i >= 0; i--) {
  
            // If arr[i] is equal to i
            if (arr[i] == i) {
                Console.Write(i);
                return;
            }
        }
  
        // If there is no such index
        Console.Write(-1);
    }
  
    // Driver code
    public static void Main(String[] args)
    {
        
        // Given Input
        int []arr = { -10, -5, 0, 3, 7 };
        int n = arr.Length;
  
        // Function Call
        findLargestIndex(arr, n);
    }
}
  
// This code is contributed by shivanisinghss2110

Javascript

// JavaScript implementation of the above approach
  
// Function to  find the maximum index
// i such that arr[i] is equal to i
function findLargestIndex( arr,  n)
    {
  
        // Traversing the array from
        // backwards
        for (var i = n - 1; i >= 0; i--) {
  
            // If arr[i] is equal to i
            if (arr[i] == i) {
                document.write(i);
                return ;
            }
        }
  
        // If there is no such index
        document.write(-1);
    }
  
    // Driver code
    // Given Input
        var arr = [ -10, -5, 0, 3, 7 ];
        var n = arr.length;
  
        // Function Call
        findLargestIndex(arr, n);
  
// This code is contributed by shivanisinghss2110

输出

3

时间复杂度：O(N)
辅助空间：O(1)

如果您希望与专家一起参加现场课程，请参阅DSA 现场工作专业课程和学生竞争性编程现场课程。