[LeetCode] 565. Array Nesting 数组嵌套

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}

解法 ：

这个题实际上是找到最大的环。要注意的一点是，一旦在前一个不同的环中被访问过，它就不会出现在当前环中，可以忽略。

The idea is to, start from every number, find circles in those index-pointer-chains, every time you find a set (a circle) mark every number as visited (-1) so that next time you won't step on it again.

Java:

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public 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;     } }

Python:

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class Solution(object):     def arrayNesting(self, nums):         """         :type nums: List[int]         :rtype: int         """         ans, step, n = 0, 0, len(nums)         seen = [False] * n         for i in range(n):             while not seen[i]:                 seen[i] = True                 i, step = nums[i], step + 1             ans = max(ans, step)             step = 0         return ans

C++:

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class Solution { public:     int arrayNesting(vector<int>& a) {         size_t maxsize = 0;         for (int i = 0; i < a.size(); i++) {             size_t 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 = max(maxsize, size);         }           return maxsize;     } };

　　

　　

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