hyunkookim 님의 블로그

23. Merge k Sorted Lists https://youtu.be/q5a5OiGbT6Q?si=Oe5XE1QtO56CiD-s # Definition for singly-linked list.# class ListNode:# def __init__(self, val=0, next=None):# self.val = val# self.next = nextclass Solution: def mergeKLists(self, lists: List[Optional[ListNode]]) -> Optional[ListNode]: # 입력으로 주어진 리스트가 비어 있거나, 리스트의 길이가 0일 경우 None 반환 if not lists or len(..

74. Search a 2D Matrix class Solution: def searchMatrix(self, matrix: List[List[int]], target: int) -> bool: ROWS = len(matrix) COLS = len(matrix[0]) l, r = 0, COLS*ROWS - 1 while l target: r = mid-1 else: return True return False https://youtu.be/Ber2pi2C0j0 class Solution: def searchMatrix(self, matrix: List[List..

21. Merge Two Sorted Lists # Definition for singly-linked list.# class ListNode:# def __init__(self, val=0, next=None):# self.val = val# self.next = nextclass Solution: def mergeTwoLists(self, list1: Optional[ListNode], list2: Optional[ListNode]) -> Optional[ListNode]: dummy = ListNode() res = dummy while list1 and list2: v1 = list1...

155. Min Stack class MinStack: def __init__(self): self.min_value = [] self.stack = [] def push(self, val: int) -> None: self.stack.append(val) if len(self.min_value) == 0: self.min_value.append(val) else: self.min_value.append(min(self.min_value[-1], val)) def pop(self) -> None: self.stack.pop() self.min_val..

20. Valid Parentheses class Solution: def isValid(self, s: str) -> bool: # 닫힌 괄호를 키로 하고, 해당하는 열린 괄호를 값으로 가지는 딕셔너리 parentheses = {')': '(', '}': '{', ']': '['} # 괄호를 추적하기 위한 스택 초기화 stack = [] # for i in range(len(s)): # if s[i] in parentheses and len(stack) >0: # if stack[-1] == parentheses[s[i]]: # stack.pop() ..

206. Reverse Linked List https://leetcode.com/problems/reverse-linked-list/description/  나도 stack 으로 풀었으나, 제대로 못푼 느낌??# Definition for singly-linked list.# class ListNode:# def __init__(self, val=0, next=None):# self.val = val# self.next = nextclass Solution: def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]: # 스텍으로 푸는 법 stack = [] ..