leetcode 565. Array Nesting 最大的环的元素数量-蒲公英云

leetcode 565. Array Nesting 最大的环的元素数量

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;
    }
};