[okyungjin] WEEK 07 Solutions

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

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+"""
+N: `s`의 길이, M: `s`에서 중복을 제외한 문자의 개수
+Time: O(N)
+Space: O(min(N,M))
+"""
+class Solution:
+    def lengthOfLongestSubstring(self, s: str) -> int:
+         # 문자의 최근 인덱스를 저장
+        char_map = {}
+
+        left = 0
+        max_len = 0
+
+        for right, char in enumerate(s):
+            if char in char_map and char_map[char] >= left:
+                left = char_map[char] + 1
+
+            char_map[char] = right
+
+            curr_len = right - left + 1
+            if curr_len > max_len:
+                max_len = curr_len
+
+        return max_len

reverse-linked-list/okyungjin.py

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+# https://leetcode.com/problems/reverse-linked-list/
+
+# Definition for singly-linked list.
+# class ListNode:
+#     def __init__(self, val=0, next=None):
+#         self.val = val
+#         self.next = next
+
+"""
+Time: O(N)
+Space O(N)
+"""
+class Solution:
+    def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]:
+        if not head:
+            return None
+
+        stack = []
+
+        while head:
+            stack.append(head)
+            head = head.next
+
+        dummy_head = ListNode()
+        curr = dummy_head
+
+        while stack:
+            curr.next = stack.pop()
+            curr = curr.next
+
+        curr.next = None
+
+        return dummy_head.next
+
+
+"""
+Time: O(N)
+Space O(1)
+"""
+class Solution:
+    def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]:
+        prev = None
+        curr = head
+
+        while curr:
+            temp = curr.next
+            curr.next = prev
+            prev, curr = curr, temp
+
+        return prev