给定从0到N的N个整数的数组arr [] ，任务是计算索引数组中的组件数。
Index array means if we are at ith index then it leads to arr[i].
The component of an index array is counted when it forms a cycle. If no cycle persists or the array contains a single element then also we consider it as a component.
For Example:
Let array arr[] = {1, 2, 0, 3}
{1, 2, 0} will form one component as starting from index 0 we reach the index 0 again as:
1 -> 2(arr[1]) -> 0(arr[2]) -> 1(arr[0])
例子：
Input: arr[] = {1, 2, 3, 5, 0, 4, 6}
Output: 2
Explanation:
Below is the traversal of the 2 components:
Component 1: Start traversal from 0, then the path of traversal is given by:
1 -> 2(arr[1]) -> 3(arr[2]) -> 5(arr[3]) -> 4(arr[5]) -> 0(arr[4]) -> 1(arr[0]).
Component 2: Only 6 is unvisited it creates one more component.
So, the total components = 2.
Input: arr[] = {1, 2, 0, 3}
Output: 2
Explanation:
Below is the traversal of the 2 components:
Component 1: Start traversal from 0, then the path of traversal is given by:
1 -> 2(arr[1]) -> 0(arr[2]) -> 1(arr[0])
Component 2: Only 3 is unvisited it creates one more component.
So, the total components = 2.
方法：想法是使用DFS遍历的概念。步骤如下：
- 从第一个未访问的索引开始，该索引将是其中具有整数0的索引。
- 在DFS遍历期间，在数组中标记已访问的元素，直到元素形成一个循环为止。
- 如果形成一个循环，则意味着我们只有一个组件，因此增加了组件数量。
- 对数组中所有未访问的索引重复上述所有步骤，并计算给定索引数组中的总数。
- 如果访问了数组的所有索引，则打印已连接组件的总数。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function for DFS traversal
void dfs(int i, int a[],
map& m)
{
// Check if not visited then
// recurr for the next index
if (m[i] == 0) {
m[i] = 1;
// DFS Traversal for next index
dfs(a[i], a, m);
}
return;
}
// Function for checking which
// indexes are remaining
int allvisited(int a[], int n,
map& m)
{
for (int i = 0; i < n; i++) {
// Marks that the ith
// index is not visited
if (m[i] == 0)
return i;
}
return -1;
}
// Function for counting components
int count(int a[], int n)
{
int c = 0;
// To mark the visited index
// during DFS Traversal
map m;
// Function call
int x = allvisited(a, n, m);
while (x != -1) {
// Count number of components
c++;
// DFS call
dfs(x, a, m);
x = allvisited(a, n, m);
}
// Print the total count of components
cout << c << endl;
}
// Driver Code
int main()
{
// Given array arr[]
int arr[] = { 1, 4, 3, 5, 0, 2, 6 };
int N = sizeof(arr) / sizeof(arr[0]);
// Function Call
count(arr, N);
return 0;
} Java
// Java program for the above approach
import java.util.*;
class Main
{
// Function for DFS traversal
static void dfs(int i, int[] a,
HashMap m)
{
// Check if not visited then
// recurr for the next index
if (!m.containsKey(i))
{
m.put(i, 1);
// DFS Traversal for next index
dfs(a[i], a, m);
}
return;
}
// Function for checking which
// indexes are remaining
static int allvisited(int[] a, int n,
HashMap m)
{
for(int i = 0; i < n; i++)
{
// Marks that the ith
// index is not visited
if (!m.containsKey(i))
return i;
}
return -1;
}
// Function for counting components
static void count(int[] a, int n)
{
int c = 0;
// To mark the visited index
// during DFS Traversal
HashMap m = new HashMap<>();
// Function call
int x = allvisited(a, n, m);
while (x != -1)
{
// Count number of components
c++;
// DFS call
dfs(x, a, m);
x = allvisited(a, n, m);
}
// Print the total count of components
System.out.print(c);
}
public static void main(String[] args)
{
// Given array arr[]
int[] arr = { 1, 4, 3, 5, 0, 2, 6 };
int N = arr.length;
// Function Call
count(arr, N);
}
}
// This code is contributed by divyesh072019 Python3
# Python3 program for the above approach
# Function for DFS traversal
def dfs(i, a, m):
# Check if not visited then
# recurr for the next index
if i in m:
if m[i] == 0:
m[i] = 1
# DFS Traversal for next index
dfs(a[i], a, m)
else:
m[i] = 1
# DFS Traversal for next index
dfs(a[i], a, m)
return
# Function for checking which
# indexes are remaining
def allvisited(a, n, m):
for i in range(n):
# Marks that the ith
# index is not visited
if i in m:
if m[i] == 0:
return i
else:
return i
return -1
# Function for counting components
def count(a, n):
c = 0
# To mark the visited index
# during DFS Traversal
m = dict()
# Function call
x = allvisited(a, n, m)
while (x != -1):
# Count number of components
c += 1
# DFS call
dfs(x, a, m)
x = allvisited(a, n, m)
# Print the total count of components
print(c)
# Driver Code
if __name__=='__main__':
# Given array arr[]
arr = [ 1, 4, 3, 5, 0, 2, 6 ]
N = len(arr)
# Function Call
count(arr, N)
# This code is contributed by rutvik_56C#
// C# program for the above approach
using System;
using System.Collections;
using System.Collections.Generic;
class GFG{
// Function for DFS traversal
static void dfs(int i, int []a,
Dictionary m)
{
// Check if not visited then
// recurr for the next index
if (!m.ContainsKey(i))
{
m[i] = 1;
// DFS Traversal for next index
dfs(a[i], a, m);
}
return;
}
// Function for checking which
// indexes are remaining
static int allvisited(int []a, int n,
Dictionary m)
{
for(int i = 0; i < n; i++)
{
// Marks that the ith
// index is not visited
if (!m.ContainsKey(i))
return i;
}
return -1;
}
// Function for counting components
static void count(int []a, int n)
{
int c = 0;
// To mark the visited index
// during DFS Traversal
Dictionary m = new Dictionary();
// Function call
int x = allvisited(a, n, m);
while (x != -1)
{
// Count number of components
c++;
// DFS call
dfs(x, a, m);
x = allvisited(a, n, m);
}
// Print the total count of components
Console.Write(c);
}
// Driver Code
public static void Main()
{
// Given array arr[]
int []arr = { 1, 4, 3, 5, 0, 2, 6 };
int N = arr.Length;
// Function Call
count(arr, N);
}
}
// This code is contributed by pratham76 3时间复杂度： O(N)
辅助空间： O(N)