LeetCode刷题——Array篇(Array Nesting)

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本文介绍了一种算法，用于寻找给定数组中最长的不重复元素链路。通过遍历数组并跟踪每个元素的链路长度，该算法能在O(N)的时间复杂度内解决问题。

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].

乍看，如果用遍历的方式去查找每个元素，并找出这个元素所对应的链路，看起来是O(N2)的时间复杂度，实则不然，亦因为经过分析，只有某个元素属于某个链路的话，那么只有遍历一次这个链路就可以了。因此实际上是个O(N)的时间复杂度。
代码如下

class Solution {
public:
    int arrayNesting(vector<int>& nums) {
        int result = 0;

        for(int i = 0; i < nums.size(); i++)
        {
            if(nums[i] != INT_MAX)
            {
                int beg = nums[i], count = 0;
                while(nums[beg] != INT_MAX)
                {
                    int temp = beg;
                    beg = nums[beg];
                    count++;
                    nums[temp] = INT_MAX;
                }
                result = max(result, count);
            }

        }
        return result;
    }
};