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[yeoju] Week 7 #938
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[yeoju] Week 7 #938
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# TC:O(n^2) SC:O(1) | ||
class Solution: | ||
def lengthOfLongestSubstring(self, s: str) -> int: | ||
max_size = 0 | ||
for start in range(len(s)): | ||
saw = set() | ||
for end in range(start, len(s)): | ||
if s[end] in saw: | ||
break | ||
else: | ||
saw.add(s[end]) | ||
max_size = max(max_size, end - start + 1) # 부분 문자열의 길이와 현재까지의 최대길이값 비교 | ||
return max_size |
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# TC:O(n * m), SC:O(n * m) | ||
class Solution: | ||
def numIslands(self, grid: List[List[str]]) -> int: | ||
cnt = 0 | ||
row = len(grid) | ||
col = len(grid[0]) | ||
def dfs(r: int, c: int) : | ||
grid[r][c] = "0" | ||
for y, x in [(r, c + 1), (r + 1, c), (r - 1, c), (r, c - 1)]: # 현재 r,c 좌표에 대해 상하좌우 탐색 | ||
if 0 <= y < row and 0 <= x < col: | ||
if grid[y][x] == "1": | ||
dfs(y, x) | ||
return | ||
|
||
for r in range(row): | ||
for c in range(col): | ||
if grid[r][c] == "1": #땅 발견 시 cnt 증가하고, 발견된 땅과 연결된 땅들을 제거 | ||
cnt += 1 | ||
dfs(r, c) | ||
return cnt |
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# TC:O(n), SC:O(1) | ||
class Solution: | ||
def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]: | ||
cur = head | ||
prv = None | ||
while cur != None: | ||
tmp = cur.next # 다음 노드의 주소를 tmp에 저장 | ||
cur.next = prv # 현재 노드가 가리키는 주소를 prv로 설정 | ||
prv = cur # prv를 현재 노드로 설정 | ||
cur = tmp # 현재 노드를 그 다음 노드로 변경 | ||
return prv |
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# TC:O(n^2), SC:O(1) | ||
class Solution: | ||
def setZeroes(self, matrix: List[List[int]]) -> None: | ||
first_row = False | ||
first_col = False | ||
|
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#첫번째 행, 열 flag | ||
for r in range(len(matrix)): | ||
if matrix[r][0] == 0: | ||
first_row = True | ||
for c in range(len(matrix[0])): | ||
if matrix[0][c] == 0: | ||
first_col = True | ||
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||
#그 이외의 행, 열 flag | ||
for r in range(1, len(matrix)): | ||
for c in range(1, len(matrix[0])): | ||
if matrix[r][c] == 0: | ||
matrix[r][0] = 0 | ||
matrix[0][c] = 0 | ||
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||
# 0으로 설정 | ||
for r in range(1, len(matrix)): | ||
for c in range(1, len(matrix[0])): | ||
if matrix[r][0] == 0 or matrix[0][c] == 0: | ||
matrix[r][c] = 0 | ||
# 첫번째 행과 열에 대해 각각 0으로 설정 | ||
if first_row: | ||
for r in range(len(matrix)): | ||
matrix[r][0] = 0 | ||
if first_col: | ||
for c in range(len(matrix[0])): | ||
matrix[0][c] = 0 |
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# TC:O(n*m), SC:O(1) | ||
# 왼쪽아래로 가는 경로는 왼쪽으로 가는 경로의 경우의 수 + 아래로 가는 경로의 수 | ||
# 가장자리로 가는 경로의 경우의 수는 모두 1 | ||
# n크기의 리스트 값을 매 번 덮어써서 계산함 | ||
class Solution: | ||
def uniquePaths(self, m: int, n: int) -> int: | ||
l = [1 for _ in range(n)] | ||
for row in range(1, m): | ||
for col in range(1, n): | ||
l[col] += l[col - 1] | ||
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||
return l[-1] |
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안녕하세요 @aa601 님,
SC 를 O(1)로 적어주셨는데, 저의 경우 최악의 케이스에서 n개가 set()에 들어갈 경우 O(n)이 되는게 아닌가 생각했었는데 혹시 O(1)로 생각하신 이유가 궁금합니다~!