LeetCode 565: Array Nesting

关注

发布于: 2020 年 05 月 07 日

DescriptionA 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
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}
﻿
问题分析按照题意，数组中的N个元素已按照其数值关系，被划分为1个或多个互不相交的“闭环”；
﻿
本题要做的，就是依次找出所有的闭环，并返回环的最大长度；
因此，可通过DFS解决该问题。其中每个节点均只有确定的一个子节点，仅需依次遍历至再次回到起点。一次遍历过程将遍历一个环上的全部节点。
遍历完所有的环，返回最大长度，即可。
实现要点避免重复遍历：使用0/1数组，记录遍历状态；
Submission
class Solution:
    def arrayNesting(self, nums: List[int]) -> int:
        n = len(nums)
        # dp[i] = 0 means node needs to be traversed
        # dp[i] = 1 means node has been traversed
        dp = [0 for i in range(n)]
        res = 0
        # do dfs
        start = 0
        while(start < n):
            if(dp[start] == 1):
                start = start + 1
                continue
            dp[start] = 1
            tmp = 1
            cur = nums[start]
            while(cur != start):
                tmp = tmp + 1
                dp[cur] = 1
                cur = nums[cur]
            res = max(res, tmp)
            start = start + 1
        return res
﻿