[LeetCode] 565. Array Nesting

565. Array Nesting

The general meaning of this question is to first select a starting position in the array, and then locate the corresponding subscript according to her value, continue until a loop occurs, and finally find the longest non-loop.

Obviously, it is necessary to calculate the beginning of each position in the array, so the first thing that comes to mind is dfs.

class Solution(object):defarrayNesting(self, nums):"""        :type nums: List[int]        :rtype: int        """ifnotnums:return0​visited = {}result = 0foridxinrange(0, len(nums)):ifidxnotinvisited:result = max(
     
        



          
             

        
        
           
                
                result , self . helper ( nums , idx , visited )) return result ​def helper ( self , nums , start , visited ): """         Recursively move to the next position until it is the same as the starting position         and record the moving Number         of :param nums:         :param start:         :param visited:         :return:         """ i = start cnt = 0 while i ! = start
         

     
        







        
        
           or cnt == 0:visited[i] = Truei = nums[i]cnt += 1returncnt
            
            
            
         

Or write it directly in a function without function recursion, which also avoids the consumption caused by function recursion;

class Solution(object):defarrayNesting(self, nums):"""        :type nums: List[int]        :rtype: int        """visited = [False] *len(nums)max_cnt = 0foriinrange(len(nums)):cnt = 1first = iifnotvisited[i]:
     
        



         
        
           
            
            
              
                next_idx = nums[i]  # 下一个位置# dfswhilefirst!= next_idx:visited[next_idx] = Truenext_idx = nums[next_idx]cnt += 1max_cnt = max(max_cnt, cnt)returnmax_cnt