565. Array Nesting 阵列嵌套

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:

1. N is an integer within the range [1, 20,000].

2. The elements of A are all distinct.

3. 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index
    for (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));

来自为知笔记(Wiz)