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27 changes: 27 additions & 0 deletions binary-tree-level-order-traversal/hu6r1s.py
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BFS 를 사용해서 풀이 진행해 주셨네요!
DFS 풀이도 도전해 보시면 좋을것 같아요.

Original file line number Diff line number Diff line change
@@ -0,0 +1,27 @@
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
from collections import deque

class Solution:
def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
if not root:
return []

queue = deque([root])
result = []
while queue:
tmp = []
for _ in range(len(queue)):
node = queue.popleft()
tmp.append(node.val)

if node.left:
queue.append(node.left)
if node.right:
queue.append(node.right)
result.append(tmp)
return result
6 changes: 6 additions & 0 deletions counting-bits/hu6r1s.py
Original file line number Diff line number Diff line change
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class Solution:
def countBits(self, n: int) -> List[int]:
ans = []
for i in range(n+1):
ans.append(bin(i)[2:].count("1"))
return ans
18 changes: 18 additions & 0 deletions graph-valid-tree/hu6r1s.py
Original file line number Diff line number Diff line change
@@ -0,0 +1,18 @@
if len(edges) != n - 1:
return False

graph = [[] for _ in range(n)]
for node, adj in edges:
graph[node].append(adj)
graph[adj].append(node)

visited = set()

def dfs(node):
visited.add(node)
for adj in graph[node]:
if adj not in visited:
dfs(adj)

dfs(0)
return len(visited) == n
12 changes: 12 additions & 0 deletions house-robber-ii/hu6r1s.py
Original file line number Diff line number Diff line change
@@ -0,0 +1,12 @@
class Solution:
def rob(self, nums: List[int]) -> int:
return max(nums[0], self.helper(nums[1:]), self.helper(nums[:-1]))


def helper(self, nums):
rob1, rob2 = 0, 0
for num in nums:
new_rob = max(rob1 + num, rob2)
rob1 = rob2
rob2 = new_rob
return rob2
30 changes: 30 additions & 0 deletions meeting-rooms-ii/hu6r1s.py
Original file line number Diff line number Diff line change
@@ -0,0 +1,30 @@
from typing import (
List,
)
from lintcode import (
Interval,
)

"""
Definition of Interval:
class Interval(object):
def __init__(self, start, end):
self.start = start
self.end = end
"""
from heapq import heappush, heappop

class Solution:
"""
@param intervals: an array of meeting time intervals
@return: the minimum number of conference rooms required
"""
def min_meeting_rooms(self, intervals: List[Interval]) -> int:
# Write your code here
intervals.sort()
ends = []
for start, end in intervals:
if ends and ends[0] <= start:
heappop(ends)
heappush(ends, end)
return len(ends)
36 changes: 36 additions & 0 deletions number-of-islands/hu6r1s.py
Original file line number Diff line number Diff line change
@@ -0,0 +1,36 @@
from collections import deque

class Solution:
"""
문제를 보니 바로 그래프 탐색이 떠올라서 bfs 알고리즘을 사용해서 구현
백준 문제에서 많이 풀어보던건데 너무 오래 되어 계속 헷갈렸음
다시 공부해야함
"""
def numIslands(self, grid: List[List[str]]) -> int:
def bfs(grid, i, j):
queue = deque()
queue.append([i, j])
grid[i][j] = "0"
while queue:
x, y = queue.popleft()
for k in range(4):
nx = x + dx[k]
ny = y + dy[k]
if nx < 0 or nx >= n or ny < 0 or ny >= m:
continue
if grid[nx][ny] == "0":
continue
grid[nx][ny] = "0"
queue.append([nx, ny])


dx = [-1, 1, 0, 0]
dy = [0, 0, -1, 1]
n, m = len(grid), len(grid[0])
cnt = 0
for i in range(n):
for j in range(m):
if grid[i][j] == "1":
bfs(grid, i, j)
cnt += 1
return cnt
33 changes: 33 additions & 0 deletions word-search-ii/hu6r1s.py
Original file line number Diff line number Diff line change
@@ -0,0 +1,33 @@
class Solution:
def findWords(self, board: List[List[str]], words: List[str]) -> List[str]:
n, m = len(board), len(board[0])
res = set()

trie = {}
for word in words:
node = trie
for ch in word:
node = node.setdefault(ch, {})
node['$'] = word

def dfs(x, y, node):
ch = board[x][y]
if ch not in node:
return
nxt = node[ch]

if '$' in nxt:
res.add(nxt['$'])

board[x][y] = "#"
for dx, dy in [(1,0), (-1,0), (0,1), (0,-1)]:
nx, ny = x + dx, y + dy
if 0 <= nx < n and 0 <= ny < m and board[nx][ny] != "#":
dfs(nx, ny, nxt)
board[x][y] = ch

for i in range(n):
for j in range(m):
dfs(i, j, trie)

return list(res)