arr[i], arr[arr[i]], .. 的最长链不重复

给定一个大小为 n 的数组，使得数组中的元素是不同的，并且范围从 0 到 n-1。我们需要找出最长链的长度 {arr[i], arr[arr[i]], arr[arr[arr[i]]]……} 使得没有集合元素重复。

例子：

Input : arr[] = [5, 4, 0, 3, 1, 6, 2]
Output :4
Explanation:
The longest chain without repetition is
{arr[0], arr[5], arr[6], arr[2]} = {5, 6, 2, 0}


Input : arr[] = {1, 0, 4, 2, 3}
Output :3
Explanation:
The longest chain without repetition is
{arr[2], arr[4], arr[3]} = {4, 2, 3}

一个天真的解决方案是从每个元素开始找到最长链的长度。要跟踪访问过的节点，请保留一个访问过的数组，并在每次从元素中找到最长链后重置此数组。

一个有效的解决方案是遍历每个索引并将它们设置为-1。一旦我们看到一个我们设置为 -1 的索引，我们就知道我们遇到了相同的索引，所以我们停下来比较当前计数是否大于我们迄今为止看到的最大值。我们可以将这个设置为 -1，这样我们就不会一次又一次地重新访问相同的索引。之所以可行，是因为每个数字都是不同的，因此总是只有一种方法可以达到特定的索引。因此，无论您在特定周期中从哪个索引开始，您总是会看到相同的周期，因此会看到相同的计数。因此，一旦一个循环被完全访问，我们就可以跳过检查这个循环中的所有索引。

C++

// C++ program to find longest chain of
// arr[i], arr[arr[i]], arr[arr[arr[i]]]
// without repetition.
#include 
using namespace std;
 
void aNesting(int arr[], int start, int& max)
{
    // Local maximum
    int c_max = 0;
 
    // if current element is not visited then
    // increase c_max by one, set arr[start]
    // and change the start to current value.
    while (arr[start] != -1) {
        c_max++;
        int temp = arr[start];
        arr[start] = -1;
        start = temp;
    }
    
   // if local max is greater than global max
   // then update global maximum
   if (c_max > max) max = c_max;
}
 
int maxLength(int arr[], int n)
{
    int max = 0;
    for (int i = 0; i < n; i++) {
        aNesting(arr, i, max);
    }
    return max;
}
 
// Driver code
int main()
{
    int arr[] = { 5, 4, 0, 3, 1, 6, 2 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << maxLength(arr, n);
    return 0;
}

Java

// Java program to find longest chain of
// arr[i], arr[arr[i]], arr[arr[arr[i]]]
// without repetition.
import java.util.*;
 
class solution
{
 
static int aNesting(int arr[], int start, int max)
{
    // Local maximum
    int c_max = 0;
 
    // if current element is not visited then
    // increase c_max by one, set arr[start]
    // and change the start to current value.
    while (arr[start] != -1)
    {
        c_max++;
        int temp = arr[start];
        arr[start] = -1;
        start = temp;
    }
     
// if local max is greater than global max
// then update global maximum
     if (c_max > max)
      max = c_max;
       
     return max;
  }
 
static int maxLength(int[] arr, int n)
{
    int max = 0,max1;
    for (int i = 0; i < n; i++)
    {
        max1 = aNesting(arr, i, max);
        if(max1>max)
         max = max1;
    }
    return max;
 }
 
// Driver code
public static void main(String args[])
{
    int arr[] = { 5, 4, 0, 3, 1, 6, 2 };
    int n = arr.length;
    System.out.println(maxLength(arr, n));
}
}
 
// Thi code is contributed by
// Surendra_Gangwar

Python3

# Python 3 program to find longest chain
# of arr[i], arr[arr[i]], arr[arr[arr[i]]]
# without repetition.
def aNesting(arr,start, max):
     
    # Local maximum
    c_max = 0
 
    # if current element is not visited then
    # increase c_max by one, set arr[start]
    # and change the start to current value.
    while (arr[start] != -1):
        c_max += 1
        temp = arr[start]
        arr[start] = -1
        start = temp
     
    # if local max is greater than global
    # max then update global maximum
    if (c_max > max):
        max = c_max
         
    return max
 
def maxLength(arr, n):
    max = 0
    for i in range(0, n, 1):
        max__ = aNesting(arr, i, max)
        if (max__>max):
            max = max__
    return max__
 
# Driver code
if __name__ =='__main__':
    arr = [5, 4, 0, 3, 1, 6, 2]
    n = len(arr)
    print(maxLength(arr, n))
     
# This code is contributed by
# Shashank_Sharma

C#

// C# program to find longest chain of
// arr[i], arr[arr[i]], arr[arr[arr[i]]]
// without repetition.
using System;
 
class GFG
{
 
static int aNesting(int[] arr, int start, int max)
{
    // Local maximum
    int c_max = 0;
 
    // if current element is not visited then
    // increase c_max by one, set arr[start]
    // and change the start to current value.
    while (arr[start] != -1)
    {
        c_max++;
        int temp = arr[start];
        arr[start] = -1;
        start = temp;
    }
     
    // if local max is greater than global max
    // then update global maximum
    if (c_max > max)
    max = c_max;
     
    return max;
}
 
static int maxLength(int[] arr, int n)
{
    int max = 0, max1;
    for (int i = 0; i < n; i++)
    {
        max1 = aNesting(arr, i, max);
        if(max1>max)
        max = max1;
    }
    return max;
}
 
// Driver code
public static void Main()
{
    int[] arr = { 5, 4, 0, 3, 1, 6, 2 };
    int n = arr.Length;
    Console.WriteLine(maxLength(arr, n));
}
}
 
// Thi code is contributed by
// Akanksha Rai

PHP

 $max)
    $max = $c_max;
}
 
function maxLength($arr, $n)
{
    $max = 0;
    for ($i = 0; $i < $n; $i++)
    {
        aNesting($arr, $i, $max);
    }
    return $max;
}
 
// Driver code
$arr = array(5, 4, 0, 3, 1, 6, 2 );
$n = sizeof($arr);
echo maxLength($arr, $n);
 
// This code is contributed
// by Akanksha Rai
?>

Javascript

 
// Javascript program to find longest chain of
// arr[i], arr[arr[i]], arr[arr[arr[i]]]
// without repetition.
function aNesting(arr, start, max)
{
     
    // Local maximum
    var c_max = 0;
 
    // If current element is not visited then
    // increase c_max by one, set arr[start]
    // and change the start to current value.
    while (arr[start] != -1)
    {
        c_max++;
        var temp = arr[start];
        arr[start] = -1;
        start = temp;
    }
     
    // If local max is greater than global
    // max then update global maximum
    if (c_max > max)
        max = c_max;
     
    return max;
}
 
function maxLength(arr, n)
{
    var max = 0, max1;
    for(var i = 0; i < n; i++)
    {
        max1 = aNesting(arr, i, max);
         
        if (max1 > max)
            max = max1;
    }
    return max;
}
 
// Driver code
var arr = [ 5, 4, 0, 3, 1, 6, 2 ];
var n = arr.length;
 
document.write(maxLength(arr, n));
 
// This code is contributed by bunnyram19
 

输出：

4

时间复杂度： O(n)
辅助空间： O(1)