LeetCode-Array Nesting

Description：
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].

题意：给定一个一维数组，长度为n，数组中所有元素的取值范围为[0,n-1]；定义集合S[i] = {A[i], A[A[i]], A[A[A[i]]], … }，元素添加到集合S中的停止条件是当插入重复元素时；根据定义，返回找到满足条件的集合S的最长长度；

解法一（超时）：很容易想到的一个方法就是利用暴力求解，从数组的下标0开始一直到下标n-1，对比每一次产生的集合，返回最长的那个；但是，这种算法的时间复杂度为O(n²)；

class Solution {
    public int arrayNesting(int[] nums) {
        int maxLen = 1;
        for (int i = 0; i < nums.length; i++) {
            int preNum = i;
            Set<Integer> table = new HashSet<>();
            while(!table.contains(nums[preNum])) {
                table.add(nums[preNum]);
                preNum = nums[preNum];
            }
            maxLen = Math.max(maxLen, table.size());
        }
        return maxLen;
    }
}

解法二：观察下面的图我们可以知道，每次一个取得的一个集合S，里面所有元素构成了一个回路，这时，不管我们访问这里面的哪个元素，最终得到的集合S都是相同的，因此，我们可以通过标记这些元素，就不必做重复的运算，最终算法的时间复杂度为O(n)；
这里写图片描述

class Solution {
    public int arrayNesting(int[] nums) {
        int maxLen = 1;
        boolean[] visited = new boolean[nums.length];
        for (int i = 0; i < nums.length; i++) {
            if (!visited[i]) {
                int preNum = i;
                Set<Integer> table = new HashSet<>();
                while(!table.contains(nums[preNum])) {
                    table.add(nums[preNum]);
                    preNum = nums[preNum];
                    visited[preNum] = true;
                }
                maxLen = Math.max(maxLen, table.size());
            }
        }
        return maxLen;
    }
}

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