[Leetcode] 565. Array Nesting 解题报告

题目：

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: 6
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:

1. N is an integer within the range [1, 20,000].
2. The elements of A are all distinct.
3. Each element of A is an integer within the range [0, N-1].

思路：

这道题目给的Example 1有误，因为其output应该是4，而不是6。

把0 - (N - 1)这N个数打乱排列在数组中，它们就会形成1 - N个嵌套回路。如果对于对于任意0 <= i <= N - 1，都有nums[i] = i，那么显然最大的嵌套回路就是1；如果对于任意0 <= i <= N - 1，都有nums[i] = (i + 1) % N，那么显然最大嵌套回路就是N。所以我们试图以每个数组元素为起点，看看能够形成多大的嵌套回路，之后将该元素置为visited（本题目中我们将该元素的值置为-1）。由于嵌套回路之间是不可能重叠相交的，所以这样的重置不会影响其余嵌套回路。

虽然算法中有两重循环，但是其时间复杂度却是O(n)，请读者分析为什么？^_^

代码：

class Solution {
public:
    int arrayNesting(vector<int>& nums) {
        int ret = 0, size = 0;
        for (int i = 0; i < nums.size(); ++i) {
            size = 0;
            for (int k = i; nums[k] >= 0; ++size) { // try to start from i
                int ak = nums[k];
                nums[k] = -1;                       // mark nums[k] as visited;
                k = ak;
            }
            ret = max(ret, size);
        }
        return ret;
    }
};