565. Array Nesting

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:

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].

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