Leetcode 565. Array Nesting

题目

链接：https://leetcode.com/problems/array-nesting/

**Level: ** Medium

Discription:
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:

• 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].

代码

class Solution{
public:
    int arrayNesting(vector<int>& nums) {
        int temp=0,ret=0;
        int b=0;
        int cache=0;
        for(int i=0;i<nums.size();i++)
        {
            b = i;
            temp = 0;
            while(nums[b]!=-1)
            {
                cache=b;
                b=nums[b];
                nums[cache]=-1;
                temp++;  
            }
             ret = max(ret,temp);    
        }
        return ret;
    }
};

思考

• 算法时间复杂度为O(n),空间复杂度为O(1)，因为数组中含有0元素，因此通过取负不方便确认是否访问，直接赋值为-1。
• 这个策略是考虑到之前访问过的回路中任意一段开始，结果序列的长度都是相等的，所以直接将已经访问过的元素标记即可。