565. Array Nesting

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: 6
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.

1. Each element of A is an integer within the range [0, N-1].

题意：

一个含有n个元素的数组包含0-n-1这些数。乱序排列，集合s[i]包含nums[i],nums[nums[i]]……这些元素，求最大的集合s。

思路：

我只会O(n^2)的(o(╥﹏╥)o）。O(n)的做法是，遍历数组，因为能够如果t=nums[i]，那么以t开始的肯定不是最长的集合。所以先遍历数组，将访问过的标记，直到访问已经标记的数，再求序列的最大值即可，因为不会重复访问已经标记的数，所以它只会遍历整个数组每个数一次，因此算法的时间复杂度是O(n)。

代码：

class Solution {
    public int arrayNesting(int[] nums) {
        int n=nums.length;
        int maxn=0;
        for(int i=0;i<n;i++)
        {
            if(nums[i]!=n)
            {
                int start=nums[i],count=0;
                while(nums[start]!=n)
                {
                    int temp=start;
                    start=nums[start];
                    count++;
                    nums[temp]=n;
                }
                maxn=Math.max(maxn,count);
            }
        }
        return maxn;
    }
}