sy1214ei 님의 블로그

https://leetcode.com/problems/longest-palindrome/description/Level : EasyTopics : Hash Table | String | Greedyclass Solution: def longestPalindrome(self, s: str) -> int: # "abccccdd" -> "dccaccd" "dccbccd" # 해쉬테이블로 각 char타입의 수를 count 해준다. # 각 문자별 저장되어있는 count가 짝수이면 -> length로 모두 사용 # count가 홀수이면 -> count-1만큼 length로 사용 char_cnt = {} ..

level : EasyTopics : Linked List | Recursionhttps://leetcode.com/problems/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() # 더미 노드 생성 ..

class Solution: def dominantIndex(self, nums: List[int]) -> int: sorted_nums = sorted(nums) if sorted_nums[-1] >= sorted_nums[-2]*2: for i in range(len(nums)): if sorted_nums[-1] == nums[i]: return i else : return -1 # Time Complexity : O(n) # Space Complexity : O(n)

class Solution: def maximumProduct(self, nums: List[int]) -> int: nums = sorted(nums) return max(nums[0]*nums[1]*nums[-1],nums[-1]*nums[-2]*nums[-3])

class Solution: # def selection_sort(self, nums): # for i in range(len(nums)): # smallest = i # for j in range(i+1, len(nums)): # if nums[j] int: # nums = self.selection_sort(nums) nums.sort() result = 0 for i in range(0,len(nums), 2): result += nums[i] return result

class Solution: def intersect(self, nums1: List[int], nums2: List[int]) -> List[int]: # arr = [] -> 겹치는 숫자 저장하는 arr arr = [] for i in range(len(nums1)): for j in range(len(nums2)): if nums1[i] == nums2[j]: arr.append(nums1[i]) nums2.pop(j) break return arr # Time Complexity : O(n^..

class Solution: def intersection(self, nums1: List[int], nums2: List[int]) -> List[int]: arr = [] nums1 = set(nums1); nums2 = set(nums2) for num in nums1: if num in nums2: arr.append(num) return arr

Code 1) Not Used Setclass Solution: def containsDuplicate(self, nums: List[int]) -> bool: for i in range(len(nums)): for j in range(i+1, len(nums)): if nums[i] == nums[j]: return True return False # Time Complexity : O(n^2) # Space Complexity : O(1) Code 2) Used Setclass Solution: def containsDuplicate(self, nums: Lis..