LeetCode565. Array Nesting [C++]

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：

1、用一个num数组，记录每个嵌套的长度。该嵌套中的每个数的长度一致。

class Solution {
public:
    int arrayNesting(vector<int>& nums) {
        int maxlen=0;
        int visited[nums.size()]={0};
        for(int i=0;i<nums.size();i++)
        {
            if(visited[nums[i]]!=0)
                continue;
            int len=0;
            int j=i;
            while(visited[j]==0)
            {
                visited[j]=1;
                j=nums[j];
                len++;
            }
            maxlen=maxlen>len?maxlen:len;
        }
        return maxlen;
    }
};

直接用本身来记录visited、最优解

class Solution {
public:
    int arrayNesting(vector<int>& nums) {
        int maxlen=0;
        for(int i=0;i<nums.size();i++)
        {
            if(nums[i]==-1)
                continue;
            int len=0;
            int j=i;
            while(nums[j]!=-1)
            {
                len++;
                int temp=nums[j];
                nums[j]=-1;
                j=temp;       
            }
            maxlen=maxlen>len?maxlen:len;
        }
        return maxlen;
    }
};