Description: Merge two sorted linked lists and return it as a sorted list. The list should be made by splicing together the nodes of the first two lists. # Definition for singly-linked list. # class ListNode: # def __init__(self, val=0, next=None): # self.val = val # self.next = next ''' [1] linear-merge) code:: def mergeTwoLists(self, l1: Optional[ListNode], l2: Optional[ListNode]) -> Optional[..

Description: Given two strings s and p, return an array of all the start indices of p's anagrams in s. You may return the answer in any order. An Anagram is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once. ''' [1] sliding window and hashmap) # optimization required idea:: pMap contains every frequency of all ..

Description: Given head, the head of a linked list, determine if the linked list has a cycle in it. There is a cycle in a linked list if there is some node in the list that can be reached again by continuously following the next pointer. Internally, pos is used to denote the index of the node that tail's next pointer is connected to. Note that pos is not passed as a parameter. Return true if the..

Description: Given the head of a singly linked list, reverse the list, and return the reversed list. # Definition for singly-linked list. # class ListNode: # def __init__(self, val=0, next=None): # self.val = val # self.next = next ''' [1] iterative) code:: def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]: prev=None cur=head next=cur while cur: next=cur.next cur.next=prev pr..

Description: Given a string s, return the longest palindromic substring in s. ''' [1] brute force) idea:: by using nested-loop. check substring is palindrome, if it is, longest substring. code:: def longestPalindrome(self, s: str) -> str: longest='' for i in range(len(s)): for j in range(i,len(s)): if self.isPalindrome(s[i:j+1]) and len(longest) str: # init dp=[[0 if i!=j else 1 for i in range(l..

Description: Given an integer n, return the number of structurally unique BST's (binary search trees) which has exactly n nodes of unique values from 1 to n. ''' [1] recursive) idea:: when n=3, then 1 and 2 and 3 can be root node 1. when 1 is a root node, left-subTree will be None, right-subTree will be made by [2,3] 2. when 2 is a root node, left-subTree will be made by [1], right-subTree will ..