565. Array Nesting.md

565. Array Nesting

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A zero-indexed array A of length N contains all integers from 0 to N-1. Find and return the longest length of set S, where S[i] = {A[i], A[A[i]], A[A[A[i]]], ... } subjected to the rule below.

Suppose the first element in S starts with the selection of element A[i] of index = i, the next element in S should be A[A[i]], and then A[A[A[i]]]… By that analogy, we stop adding right before a duplicate element occurs in S.

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

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Solution

• mine

• the most votes

class Solution {
    public int arrayNesting(int[] a) {
        int maxsize = 0;
        for (int i = 0; i < a.length; i++) {
            int size = 0;
            for (int k = i; a[k] >= 0; size++) {
                int ak = a[k];
                a[k] = -1; // mark a[k] as visited;
                k = ak;
            }
            maxsize = Integer.max(maxsize, size);
        }
        return maxsize;
    }
}

• other

// O(n)time O(1)space
class Solution {
    public int arrayNesting(int[] nums) {
        int res = 0;
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] != Integer.MAX_VALUE) {
                int start = nums[i], count = 0;
                while (nums[start] != Integer.MAX_VALUE) {
                    int temp = start;
                    start = nums[start];
                    count++;
                    nums[temp] = Integer.MAX_VALUE;
                }
                res = Math.max(res, count);
            }
        }
        return res;
    }
}

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