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.

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

想到的brute force解法，但是肯定会超时的。

这道题利用的一个思路是，用一个数组记录已经访问过的节点。有一个已知条件需要用上，就是每个元素都不相同，所以才可以使用这个办法。在进行循环的时候，你不需要担心已经访问过的点会不会被其他点访问到，因为只有一种情况可以访问到该点。利用这个条件，就可以在遍历的时候，把访问过的下标设置成true，在外层循环的时候跳过这些访问过的点（因为已经访问过这个点，所以肯定比由它开始的距离要远）。

代码：

public int arrayNesting(int[] nums) {        if(nums == null || nums.length == 0) return 0;        boolean[] visited = new boolean[nums.length];        int maxCount = 0;        for(int i=0;i<nums.length;i++) {            if(visited[i] != true) {                int curCount = 0;                int curIndex = i;                do {                    curCount++;                    curIndex = nums[curIndex];                    visited[curIndex] = true;                } while(curIndex != i);                maxCount = Math.max(maxCount, curCount);            }        }        return maxCount;    }