# 565. Array Nesting 数组嵌套

@TOC

## # 题目描述

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].

## # 解题方法

``````A = [5,4,0,3,1,6,2]
One of the longest S[K]:
S[0] = {A[0], A[5], A[6], A[2]} = {5, 6, 2, 0}
``````

``````class Solution(object):
def arrayNesting(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
# 放在这里
visited = [False] * len(nums)
ans = 0
for i in xrange(len(nums)):
while not visited[i]:
# 下面两行的顺序不能变
visited[i] = True
i = nums[i]
return ans
``````

C++代码如下：

``````class Solution {
public:
int arrayNesting(vector<int>& nums) {
int res = 0;
const int N = nums.size();
vector<bool> visited(N, false);
for (int i = 0; i < N; i ++) {
int path = 0;
while (!visited[i]) {
visited[i] = true;
path += 1;
i = nums[i];
}
res = max(res, path);
}
return res;
}
};
``````

## # 日期

2018 年 3 月 6 日 2018 年 12 月 15 日 —— 今天四六级