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]. 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; }