206. Reverse Linked List

Given the head of a singly linked list, reverse the list, and return the reversed list.


Note: See prev is returned.



def reverseList(self, head: Optional[ListNode]):
   
    cur = head
    prev = None
   
    while cur:            
        origNext = cur.next            
        cur.next = prev                                    
        prev = cur
        cur = origNext
   

    return prev 

Comments

Post a Comment

Popular posts from this blog

849. Maximize Distance to Closest Person

You are given an array representing a row of  seats  where  seats[i] = 1  represents a person sitting in the  i th  seat, and  seats[i] = 0  represents that the  i th  seat is empty  (0-indexed) . There is at least one empty seat, and at least one person sitting. Alex wants to sit in the seat such that the distance between him and the closest person to him is maximized.  Return  that maximum distance to the closest person . def maxDistToClosest ( self , seats : List [ int ]):                     taken = []         for i , occupied in enumerate ( seats ):         if occupied :             taken . append ( i )         S = len ( seats )     T = len ( taken )                 maxDist = max ( taken [ 0 ], S - taken [- 1 ] - 1 ) ...
Read more

347. Top K Frequent Elements

Given an integer array  nums  and an integer  k , return  the   k   most frequent elements . You may return the answer in  any order .   Example: Input: nums = [1,1,1,2,2,3], k = 2 Output: [1,2] def topKFrequent ( self , nums : List [ int ], k : int ):             freq = Counter ( nums )         return heapq . nlargest ( k , freq . keys (), key = freq . get )    
Read more

139. Word Break

Given a string  s  and a dictionary of strings  wordDict , return  true  if  s  can be segmented into a space-separated sequence of one or more dictionary words. Note  that the same word in the dictionary may be reused multiple times in the segmentation.   Example: Input: s = "leetcode", wordDict = ["leet","code"] Output: true Explanation: Return true because "leetcode" can be segmented as "leet code".   def wordBreak ( s : str , wordDict : List [ str ]):             word_set = set ( wordDict )     dp = [ False ] * ( len ( s ) + 1 )     dp [ 0 ] = True     for i in range ( 1 , len ( s ) + 1 ):         for j in range ( i ):             if dp [ j ] and s [ j : i ] in word_set :                 dp [ i ] = True                 break   ...
Read more