565. Array Nesting

Posted 2020-10-19 唐僧洗发爱飘柔

　　　　这道题为中等题

　　题目:

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: 6
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].

 

　　思路：

　　　　这道题思路还是比较简单，最开始我没仔细看题目要求，他的列表中没有重复项，结果导致代码超时。这个题主要用另外一个列表来判断这个元素是否被遍历，如果被搜索过就将其值置为1，否则计数值num+=1，每遍历一个元素就比较一次max_num.

　　代码：

class Solution(object):
    def arrayNesting(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        
        n = len(nums)
        a = [0 for j in range(n)]
        num = 0
        max_num = 0
        for i in range(n):
            num = 0
            while a[i] != 1:
                a[i] = 1
                i = nums[i]
                num += 1
            max_num = max(max_num, num)   
        return max_num