Merge pull request #1475 from sejineer/main

[sejineer] Week 07 solutions

Author: SamTheKorean

longest-substring-without-repeating-characters/sejineer.py

@@ -0,0 +1,21 @@
+"""
+시간 복잡도: O(N)
+공간 복잡도: O(N) 
+"""
+class Solution:
+    def lengthOfLongestSubstring(self, s: str) -> int:
+        s_set = set()
+        start, end = 0, 0
+
+        result = 0
+
+        while end < len(s):
+            if s[end] in s_set:
+                s_set.remove(s[start])
+                start += 1
+            else:
+                s_set.add(s[end])
+                end += 1
+                result = max(result, end - start)
+        
+        return result

number-of-islands/sejineer.py

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+"""
+시간 복잡도: O(m * n)
+공간 복잡도: O(m * n)
+"""
+from collections import deque
+
+class Solution:
+    def numIslands(self, grid: List[List[str]]) -> int:
+        m, n = len(grid), len(grid[0])
+        vis = [[False] * n for _ in range(m)]
+
+        dx = [1, -1, 0, 0]
+        dy = [0, 0, 1, -1]
+
+        result = 0
+        for i in range(m):
+            for j in range(n):
+                if grid[i][j] == '1' and not vis[i][j]:
+                    queue = deque()
+                    queue.append((j, i))
+                    vis[i][j] = True
+                    while queue:
+                        cur = queue.popleft()
+                        for nxt in range(4):
+                            nx = cur[0] + dx[nxt]
+                            ny = cur[1] + dy[nxt]
+                            if not 0 <= nx < n or not 0 <= ny < m:
+                                continue
+                            if vis[ny][nx] or grid[ny][nx] != '1':
+                                continue
+                            queue.append((nx, ny))
+                            vis[ny][nx] = True
+                    result += 1
+        
+        return result

reverse-linked-list/sejineer.py

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+"""
+시간 복잡도: O(N)
+공간 복잡도: O(1)
+"""
+class Solution:
+    def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]:
+        prev = None
+
+        while head:
+            nxt = head.next
+            head.next = prev
+            prev = head
+            head = nxt
+        
+        return prev