Array Nesting 解答

Question

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

这道题的关键是建立visited数组，如果起始的元素已经被遍历过了，那么我们就不用再遍历。注意：因为Note里强调了Input数组中没有相同的元素，我们只需要检查起始元素有无被遍历。

class Solution:
    def arrayNesting(self, nums: List[int]) -> int:
        result = -1
        n = len(nums)
        visited = [False] * n
        for i in range(n):
            if visited[i]:
                continue
            count, j = 0, i
            while count == 0 or j != i:
                visited[j] = True
                j = nums[j]
                count += 1
                result = max(result, count)
        return result