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].

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題目描述：求一個數組中的連續嵌套的子數組的最長大小
（這道題看了好幾遍才知道求什麼，還有這裏的output應該是輸出爲4，但是這裏寫的是6）

思路：看到是求數組的嵌套問題，最初的思路是複製一個數組，然後在數組中進行查找，如果找到後進行標記，防止出現P形的循環

後來看到討論區的答案，發現不需要複製，直接可以在原來的數組中進行標記,
原因是題中的注意，N 的範圍是[1 ,20000], A 的範圍是[0, N-1] 而且數組沒有重複項，所以最長子數組的形狀只可能是一個或者多個O的環，所以不必擔心標記後改變了最長子數組中的元素值

public int arrayNesting(int[] a) {
        int size = 0,maxsize = 0;
        for (int i = 0; i < a.length; i++) {
            size = 0;
            for (int k = i; a[k] >= 0; size++){
                int va = a[k];
                a[k] = -1;
                k = va;
            }
            maxsize = Math.max(size,maxsize);
        }
       return maxsize;