[leetcode-565-Array Nesting]

A zero-indexed array A consisting of N different integers is given. The array contains all integers in the range [0, N - 1].

Sets S[K] for 0 <= K < N are defined as follows:

S[K] = { A[K], A[A[K]], A[A[A[K]]], ... }.

Sets S[K] are finite for each K and should NOT contain duplicates.

Write a function that given an array A consisting of N integers, return the size of the largest set S[K] for this array.

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:

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

思路：

感觉类似于求集合的概念。根据给出的例子，可以发现，5 6 2 0这四个数字无论是从0开始还是从2开始，始终是这四个数字为一个集合，于是就可以用一个标记用来表示

是否已经遍历过该数字，比如从0开始，依次找到A[0], A[5], A[6], A[2]，将他们依次做标记，就可以避免重复遍历。

int arrayNesting(vector<int>& nums)
{
  int n = nums.size();
  vector<int>flags(n,false);
  int res =0;
  int num =0;
  for(int i =0;i<n;i++)
  {
    if(flags[i] == true)continue;
    num = 0;
    for(int j = i;flags[j]==false;)
    {
      num++;
      flags[j] = true;
      j = nums[j];
    }
    res = max(res,num);     
  }
  return res;
 }