動機

0~n就是loop!!

Problem

You are given an integer array nums of length n where nums is a permutation of the numbers in the range [0, n - 1].

You should build a set s[k] = {nums[k], nums[nums[k]], nums[nums[nums[k]]], ... } subjected to the following rule:

  • The first element in s[k] starts with the selection of the element nums[k] of index = k.
  • The next element in s[k] should be nums[nums[k]], and then nums[nums[nums[k]]], and so on.
  • We stop adding right before a duplicate element occurs in s[k].

Return the longest length of a set s[k].

 

Example 1:

Input: nums = [5,4,0,3,1,6,2]Output: 4Explanation: nums[0] = 5, nums[1] = 4, nums[2] = 0, nums[3] = 3, nums[4] = 1, nums[5] = 6, nums[6] = 2.One of the longest sets s[k]:s[0] = {nums[0], nums[5], nums[6], nums[2]} = {5, 6, 2, 0}

Example 2:

Input: nums = [0,1,2]Output: 1

 

Constraints:

  • 1 <= nums.length <= 105
  • 0 <= nums[i] < nums.length
  • All the values of nums are unique.

Sol

只要是排序就是會有各種loop，所以用dfs去求loop長度就好

class Solution:
    def arrayNesting(self, nums: List[int]) -> int:
        def dfs(i):
            if i == -1:
                return -1
            else:
                j, nums[i] = nums[i], -1
                return 1+dfs(j)

        return max([dfs(i) for i in range(len(nums))])