https://leetcode.cn/problems/array-nesting/
首先想到的是直接暴力遍历,以数组中每一个元素为起点，看最远能走多远：
from typing import List
class Solution:
# 暴力
def arrayNesting(self, nums: List[int]) -> int:
result = 0
for x in nums:
next_n = x
s = set()
while next_n not in s:
s.add(next_n)
next_n = nums[next_n]
if len(s) > result:
result = len(s)
return result
一顿操作，直接超时。
思考一下，直接遍历会有很多重复的操作。 如果某个元素已经出现过了，那么就不用再看它了。
from typing import List
class Solution:
# 暴力
def arrayNesting(self, nums: List[int]) -> int:
has_appeared = set()
result = 0
for x in nums:
if x in has_appeared:
continue
next_n = x
s = set()
while next_n not in s:
has_appeared.add(next_n)
s.add(next_n)
next_n = nums[next_n]
if len(s) > result:
result = len(s)
return result
