題目

鏈接：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].

代碼

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

• 算法時間復雜度為O(n),空間復雜度為O(1)，因為數組中含有0元素，因此通過取負不方便確認是否訪問，直接賦值為-1。
• 這個策略是考慮到之前訪問過的回路中任意一段開始，結果序列的長度都是相等的，所以直接將已經訪問過的元素標記即可。

轉載于:https://www.cnblogs.com/zuotongbin/p/10219981.html

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