leetcode 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].

本题题意有点绕，本质上就是就求环的元素的数量，直接暴力求解即可，注意使用标志数组来处理

代码如下：

#include <iostream>#include <vector>#include <map>#include <set>#include <queue>#include <stack>#include <string>#include <climits>#include <algorithm>#include <sstream>#include <functional>#include <bitset>#include <numeric>#include <cmath>using namespace std;class Solution {public:    int arrayNesting(vector<int>& nums)     {        int maxLen = 0;        vector<bool> visit(nums.size(), false);        for (int i = 0; i < nums.size(); i++)        {            if (visit[i] == false)            {                int j = i, count = 0;                while (count == 0 || j != i)                {                    visit[j] = true;                    j = nums[j];                    count++;                }                maxLen = max(maxLen,count);            }        }        return maxLen;    }};