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: 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].

Solution1：DFS 尝试

思路：
Time Complexity: O(N) Space Complexity: O(N)

Solution2：简洁版

思路：
Time Complexity: O(N) Space Complexity: O(N)

Solution1 Code：

public class Solution {
    public int arrayNesting(int[] nums) {
        int max = Integer.MIN_VALUE;
        boolean[] visited = new boolean[nums.length];
        
        for (int i = 0; i < nums.length; i++) {
            if (visited[i]) continue;
            max = Math.max(max, calcLength(nums, i, visited));
        }
        return max;
    }
    
    private int calcLength(int[] nums, int start, boolean[] visited) {
        int i = start, count = 0;
        while (count == 0 || i != start) {
            visited[i] = true;
            i = nums[i];
            count++;
        }
        return count;
    }
}

Solution2 Code：

public class Solution {
    public int arrayNesting(int[] a) {
        int maxsize = 0;
        for (int i = 0; i < a.length; i++) {
            int size = 0;
            for (int k = i; a[k] >= 0; size++) {
                int ak = a[k];
                a[k] = -1; // mark a[k] as visited;
                k = ak;
            }
            maxsize = Integer.max(maxsize, size);
        }

        return maxsize;
    }
}