Array Nesting

题目描述:

A zero-indexed array A consisting of N different integers is given. The array contains all integers in the range [0, N - 1].

Sets S[K] for 0 <= K < N are defined as follows:

S[K] = { A[K], A[A[K]], A[A[A[K]]], ... }.

Sets S[K] are finite for each K and should NOT contain duplicates.

Write a function that given an array A consisting of N integers, return the size of the largest set S[K] for this array.

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 array A is an integer within the range [0, N-1].

题目大意:

给定一个数组nums。题目要我们求出新的一个数组S。求S的规则就是：S[K] = { A[K], A[A[K]], A[A[A[K]]], ... }.最后返回数组S中最大的元素即可。
我们可以直接模拟，提交后发现运行超时。
手动模拟一下给出的示例：

A = [5,4,0,3,1,6,2]S[0]={5,6,2,0};
S[1]={4,1};
S[2]={0,5,2,6};
S[3]={3};
S[4]={1,4};
S[5]={6,2,0,5};
S[6]={2,0,5,6};

我们发现，其实在算出S[0]的长度后，计算S[2],S[5],S[6]都重复了操作。因为nums数组中的数字是在0～n-1之间的（n是nums数组的长度）。所以每一个S元素都会形成一个环。也就是说，如果当前的元素在之前的环中出现过，那么就不必再计算以该元素开头的S序列的长度了。这样势必会减少大量的操作。
这里用vis数组初始化为0表示元素没有在之前的环中出现过，等于1表示已经出现过在之前的环中。

题目代码:

class Solution {
public:int arrayNesting(vector<int>& nums) {int maxSize = INT_MIN;int size = 0;int vis[nums.size()] = {0};for(int i = 0; i < nums.size(); i++){size = 0;int next = i;while(!vis[next]){vis[next] = 1;size++;next = nums[next];}maxSize = max(maxSize, size);}return maxSize;}
};

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Array Nesting

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