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概述

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}
class Solution {
public:
    int arrayNesting(vector<int>& nums) {
        int n = nums.size();
        vector<bool>visited(n);
        int max_len = 1;
        for(int i = 0;i<n;i++)
        {
            if(visited[i]) continue;
            int next = nums[i];visited[i] = true;
            int len = 1;
            while(!visited[next])
            {
                visited[next] = true;
                next = nums[next];
                len++;
            }
            max_len = max(max_len,len);
        }
        return max_len;
    }
};

 

转载于:https://www.cnblogs.com/jxr041100/p/8092902.html

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