Skip to content
Merged
Show file tree
Hide file tree
Changes from 4 commits
Commits
File filter

Filter by extension

Filter by extension

Conversations
Failed to load comments.
Loading
Jump to
Jump to file
Failed to load files.
Loading
Diff view
Diff view
34 changes: 34 additions & 0 deletions binary-tree-level-order-traversal/pmjuu.py
Copy link
Contributor

Choose a reason for hiding this comment

The reason will be displayed to describe this comment to others. Learn more.

Queue 를 활용한 BFS 방식으로 문제를 풀어주셨네요!
DFS를 사용하여 풀어보시는것도 재미있을것 같습니다 :)

Original file line number Diff line number Diff line change
@@ -0,0 +1,34 @@
'''
시간 복잡도: O(n)
공간 복잡도: O(n)
'''

from collections import deque
from typing import List, Optional

class TreeNode:
def __init__(self, val=0, left=None, right=None):
self.val = val
self.left = left
self.right = right

class Solution:
def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
queue = deque([root]) if root else []
result = []

while queue:
same_level_nodes = []

for _ in range(len(queue)):
node = queue.popleft()
same_level_nodes.append(node.val)

if node.left:
queue.append(node.left)
if node.right:
queue.append(node.right)

result.append(same_level_nodes)

return result
15 changes: 15 additions & 0 deletions counting-bits/pmjuu.py
Original file line number Diff line number Diff line change
@@ -0,0 +1,15 @@
'''
시간 복잡도: O(n)
공간 복잡도: O(n)
'''
from typing import List


class Solution:
def countBits(self, n: int) -> List[int]:
dp = [0] * (n + 1)

for i in range(1, n + 1):
dp[i] = dp[i >> 1] + (i & 1)

return dp
25 changes: 25 additions & 0 deletions house-robber-ii/pmjuu.py
Original file line number Diff line number Diff line change
@@ -0,0 +1,25 @@
'''
시간 복잡도: O(n)
공간 복잡도: O(n)
'''
from typing import List

class Solution:
def rob(self, nums: List[int]) -> int:
n = len(nums)
if n == 1:
return nums[0]
if n == 2:
return max(nums[0], nums[1])

dp_first = [0] * n
dp_second = [0] * n

dp_first[0], dp_first[1] = nums[0], max(nums[0], nums[1])
dp_second[1], dp_second[2] = nums[1], max(nums[1], nums[2])

for i in range(2, n):
dp_first[i] = max(dp_first[i - 1], dp_first[i - 2] + nums[i])
dp_second[i] = max(dp_second[i - 1], dp_second[i - 2] + nums[i])

return max(dp_first[-2], dp_second[-1])