Leetcode——565. Array Nesting

题目原址

https://leetcode.com/problems/array-nesting/description/

题目描述

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：

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：

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].

解题思路

给一个长度N的数组，里面的元素值为 1 到 N-1，要求找到最长的子数组，返回其长度。
子数组其实是嵌套的数组，题目中解释了什么是嵌套数组，即元素值变为坐标，再讲坐标变为值。

因为数组中可能存在多个循环子数组，那就一一找到，最后返回一个长度最大的即可。

每个元素都要找循环子数组。将元素存入set集合中，如果集合中不存在该元素，说明没有构成循环，则将元素存到set集合中，并将下标更改为当前元素的值。将循环子串长度的计数元素值+1。
每个元素都如上述步骤进行计算，最后找到数组中最长的子数组的长度。

AC代码

class Solution {Set<Integer> set = new HashSet<Integer>();public int arrayNesting(int[] nums) {int ret = 0;for(int i: nums) {ret = Math.max(ret, valid(nums,i));}return ret;}//返回循环的子串中元素的个数public int valid(int[] nums,int i) {int ret = 0;while(!set.contains(i)) {set.add(i);i = nums[i];ret++;}return ret;}
}