# 565 Array Nesting
###### tags: `leetcode`
## Problem Statement
You are given an integer array nums of length n where nums is a permutation of the numbers in the range $[0, n - 1]$.

You should build a set ```s[k] = {nums[k], nums[nums[k]], nums[nums[nums[k]]], ... }``` subjected to the following rule:

- The first element in s[k] starts with the selection of the element nums[k] of index = k.
- The next element in s[k] should be nums[nums[k]], and then nums[nums[nums[k]]], and so on.

We stop adding right before a duplicate element occurs in s[k].
Return the longest length of a set s[k].

- Example 1:

> Input: nums = [5,4,0,3,1,6,2]
Output: 4
Explanation:
nums[0] = 5, nums[1] = 4, nums[2] = 0, nums[3] = 3, nums[4] = 1, nums[5] = 6, nums[6] = 2.
One of the longest sets s[k]:
s[0] = {nums[0], nums[5], nums[6], nums[2]} = {5, 6, 2, 0}
- Example 2:

> Input: nums = [0,1,2]
Output: 1

- Constraints:

> $1 \leq nums.length \leq 10^5\\
0 \leq nums[i] < nums.length$
All the values of nums are unique
## Solution
- Because the number would not be overlapped, use an array to see whether the number has been used.

 vector<int> check(nums.size(), 0);

- Check the status and keep on adding until the number has been used.

ans= 0;
while (check[idx]== 0)
{
    check[idx]= 1;
    ans++;
    idx= nums[idx];
}

- Two things for transition, first is to see whether or not to replace the maximum number, second is to look for the next clean number.

if (_max< ans)
    _max= ans;
while (check.at(i))
{
    if (++i== nums.size())
        return _max;
}
idx= i;