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:

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

Solution

(1) Java

(2) Python

class Solution:
    def arrayNesting(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        rst = 0
        for i, v in enumerate(nums):
            tmp = 0
            j = i
            while v >= 0:
                nums[j]  = -v if v != 0 else -len(nums)
                tmp += 1
                j = v
                v = nums[v]
            rst = max(rst, tmp)
        return rst

(3) Scala