565. Array Nesting

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: 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:

1. N is an integer within the range [1, 20,000].
2. The elements of A are all distinct.
3. Each element of array A is an integer within the range [0, N-1].

这题很显然是要找数组中的所有闭环。从下标0开始遍历，凡是在访问过的数组都标-1以示访问过。并维护一个max变量以记录最大的闭环长度。

自己一开始的渣代码如下：

class Solution {public:    int arrayNesting(vector<int>& nums) {        vector<int> cycle(nums.size() + 1, 0);        int len = 0;        for(int i = 0; i < nums.size(); ++ i){            if(cycle[nums[i]] != 0) continue;            findCycleLength(nums, cycle, nums[i]);        }        int max = INT_MIN;        for(auto len : cycle){            if(max < len) max = len;        }                return max;    }        void findCycleLength(vector<int>& nums, vector<int>& cycle, int cur){        vector<int> route;        route.push_back(cur);        int curLen = 1;        while(nums[cur] != route[0]){            curLen++;            cur = nums[cur];            route.push_back(cur);        }        for(auto idx : route){            cycle[idx] = curLen;        }    }};

参考大腿的代码：

class Solution {public:    int arrayNesting(vector<int>& nums) {        int maxx = 0;        for(int i = 0; i < nums.size(); ++i){            int len = 0;            for(int k = i; nums[k] >= 0; len ++){                int ak = nums[k];                nums[k] = -1;                k = ak;            }            maxx = max(maxx, len);        }        return maxx;    }};