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: 4
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:

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].

解题思路：

有点类似并查集,对于一个长度实际上肯定是形成一个封闭的环的，这里用递归统计了长度。速度有点慢。

class Solution {
private:
    vector<int> know;
    int count=0;
    unordered_map<int,int> exist;
    int deep(vector<int>& nums,int i){
        if(exist.count(i) > 0) {return count;}

            exist[i]++;
            count++;
            know[i] = deep(nums,nums[i]);
            return know[i];

    }

public:
    int arrayNesting(vector<int>& nums) {
        int max_digit=0;
        for(int i=0;i<nums.size();i++){
            know.push_back(-1);
        }
        vector<int> store;
        for(int i=0;i<nums.size();i++){
            exist.clear();
            count=0;
            if(know[i] == -1)
                know[i] = deep(nums,i);

            if(know[i]>max_digit) max_digit = know[i];


        }
        return max_digit;

    }
};