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: 6Explanation: 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].
/*** @param {number[]} nums* @return {number}*/var arrayNesting = function (nums) {let res = 0, set = new Set(); //因为集合S会成环，因此可以用set存储遍历过的indexfor (let i = 0; i < nums.length; i++) {let curIndex = i, count = 0;while (!set.has(curIndex)) {set.add(curIndex);curIndex = nums[curIndex];count++;}res = Math.max(res, count);}return res;};//let arr = [5, 4, 0, 3, 1, 6, 2];let arr = [1, 2, 3, 4, 5, 0]console.log(arrayNesting(arr));