[leetcode] 565. Array Nesting

[leetcode] 565. Array Nesting

Description

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: 4Explanation: 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].

分析

其实对于遍历过的数字，我们不用再将其当作开头来计算了，而是只对于未遍历过的数字当作嵌套数组的开头数字，不过在进行嵌套运算的时候，并不考虑中间的数字是否已经访问过，而是只要找到和起始位置相同的数字位置，然后更新结果res。

代码

class Solution {public: int arrayNesting(vector& nums) { int n=nums.size(); int res=0; vector visited(n,false); for(int i=0;i& nums,int start, vector& visited){ int i=start; int cnt=0; while(cnt==0||i!=start){ visited[i]=true; i=nums[i]; ++cnt; } return cnt; }};

参考文献

​​[LeetCode] Array Nesting 数组嵌套​​

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