标签：from hat sele selection art star cte index ext
565. Array Nesting
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:
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; k=ak; } maxsize=max(maxsize,size); } return maxsize; } };
python代码
class Solution(object): def arrayNesting(self, nums): """ :type nums: List[int] :rtype: int """ max_sz=0 for i in xrange(len(nums)): k=i sz=0 while nums[k]>=0: nk=nums[k] nums[k]=-1 k=nk sz+=1 #print "sz: ",sz max_sz=max(max_sz,sz) #print "nums: ",nums return max_sz
标签：from hat sele selection art star cte index ext
原文地址：https://www.cnblogs.com/learning-c/p/9281193.html
