leetcode-565. Array Nesting

565. Array Nesting

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].

题意

给定一个数组 a，求一个集合 S 的长度，要求 S 中的元素满足：
s[0] = a[i], s[1] = a[a[i]], s[2] = a[a[a[i]]] ....

思路

最容易想到的就是遍历原数组，以a[i]为起点，求出每个数组的长度，取最长的返回。用一个数组来记录是否遍历过该结点即可
一开始想到会不会TLE，但是结果并没有。。

class Solution {    public int arrayNesting(int[] nums) {        int[] visited = new int[nums.length];        int maxlen = 0;        for(int i = 0;i<nums.length;i++){            if (visited[i] == 0) {                visited[i] = 1;                int count = 1;                int index = nums[i];                while (visited[index]==0) {                    visited[index] = 1;                    index = nums[index];                    ++count;                }                maxlen = Math.max(count,maxlen);            }        }        return maxlen;    }}