[leetCode]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].

我的方法：用了递归，结果超时 

public class Solution {    //0  2 1    public int count(int[] nums,int index, int cur){        if(nums[nums[cur]]==index) return 1;        return 1+count(nums, index, nums[cur]);    }        public int arrayNesting(int[] nums) {        int[] res = new int[nums.length];        int max;        for(int i=0; i<res.length; i++){            res[i]=1;           }                for(int i=0; i<nums.length; i++){            if(nums[i]==i) continue;            else{                res[i]=1+count(nums, i, i);            }        }                max=res[0];        for(int i=1; i<res.length;i++){            if(res[i]>max)                max=res[i];        }                return max;    }}

应该要选用合适的数据结构来辅助算法。HashSet

public class Solution {    HashSet<Integer> set = new HashSet();    public int arrayNesting(int[] nums) {        int res=0;        for(int i=0; i<nums.length; i++){            res=Math.max(res,check(nums[i],nums));        }                return res;    }        public int check(int k, int[] nums){        int count=0;        while(!set.contains(k)){            count++;            set.add(k);            k=nums[k];        }                return count;    }}

这里的HashSet里面最终会装满nums里面全部的值。注意check函数里面最开始是把count初始化为0的。

还有，注意这两个方法中存储最大值max的方法，后者比较好。

方法三

int arrayNesting(vector<int>& nums) {        if(nums.empty()) return 0;        int ret_size = 0;        for(int i=0; i<nums.size(); i++){            if(nums[i] < 0) continue;            int cur = nums[i];            nums[i] = -nums[i];            set<int> st;            while(!st.count(cur) && cur >= 0){                st.insert(cur);                int temp = nums[cur];                nums[cur] = -nums[cur];                cur = temp;            }            if(st.size() > ret_size) ret_size = st.size();        }        return ret_size;    }