題目：

A zero-indexed array A of length N contains all integers from 0 to N-1. Find and return the longest length of set S, where S[i] = {A[i], A[A[i]], A[A[A[i]]], … } subjected to the rule below.
Suppose the first element in S starts with the selection of element A[i] of index = i, the next element in S

should be A[A[i]], and then A[A[A[i]]]… By that analogy, we stop adding right before a duplicate element occurs in S.
Example 1:

Input: A = [5,4,0,3,1,6,2] 
Output: 4 
Explanation:  A[0] =5, A[1] = 4, A[2] = 0, A[3] = 3, A[4] = 1, A[5] = 6, A[6] = 2.
One of the longest S[K]: S[0] = {A[0], A[5], A[6], A[2]} = {5, 6, 2, 0} 
 

Note:
N is an integer within the range [1, 20,000].
The elements of A are all distinct.
Each element of A is an integer within the range [0, N-1].

解釋：
注意，開始出現迴圈的時候就結束，結束的條件就是next==i。
python程式碼：

class Solution(object):
    def arrayNesting(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
 

        _max=0
        list_set=set()
        for i, num in enumerate(nums):
            if num not in list_set:
                list_set.add(num)
                count=1
                next=num
                #如果滿足條件則會回到原點
                while next!=i:
                    next=nums[next]
                    list_set.add(next)
                    count+=1
                _max=max(_max,count)
        return _max

c++程式碼：

#include <set>
using namespace std;
class Solution {
public:
    int arrayNesting(vector<int>& nums) {
        int maxLen=1;
        int n=nums.size();
        vector<bool> visited(n,false);
        for (int i=0;i<n;i++)
        {
            int  next=nums[i];
            int _count=1;
            if(!visited[next])
            {
                visited[next]=true;
                while (next!=i)
                {
                    next=nums[next];
                    visited[next]=true;
                    _count++;
                }
                maxLen=max(maxLen,_count);
            }
        }
        return maxLen;
    }
};

總結：
發現了一個規律，就是c++用vector<bool>作為visited比用set<int>作為visited速度更快，好幾次用set都超時了。