feat(work): add LeetCode solutions for problems 380, 82, 92

work/1/Medium/Linked List/82.remove-duplicates-from-sorted-list-ii.py

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+"""
+82. Remove Duplicates from Sorted List II
+Difficulty: Medium
+https://leetcode.com/problems/remove-duplicates-from-sorted-list-ii/
+
+──────────────────────────────────────────────────
+
+You are given the head of a sorted linked list.
+
+Delete all nodes that have duplicate numbers, leaving only distinct
+numbers from the original list.
+
+Return the linked list sorted as well.
+
+
+
+Example 1:
+
+Input: head = [1,2,3,3,4,4,5]
+Output: [1,2,5]
+
+Example 2:
+
+Input: head = [1,1,1,2,3]
+Output: [2,3]
+
+
+
+Constraints:
+
+        • The number of nodes in the list is in the range [0, 300].
+
+        • -100 <= Node.val <= 100
+
+        • The list is guaranteed to be sorted in ascending order.
+"""
+
+
+# Definition for singly-linked list.
+# class ListNode:
+#     def __init__(self, val=0, next=None):
+#         self.val = val
+#         self.next = next
+class Solution:
+    def deleteDuplicates(self, head: ListNode | None) -> ListNode | None:
+        sentinel = ListNode(0, head)
+        pred = sentinel
+
+        while head:
+            if head.next and head.val == head.next.val:
+                while head.next and head.val == head.next.val:
+                    head = head.next
+
+                pred.next = head.next
+            else:
+                pred = pred.next
+
+            head = head.next
+
+        return sentinel.next

work/1/Medium/Linked List/92.reverse-linked-list-ii.py

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+"""
+92. Reverse Linked List II
+Difficulty: Medium
+https://leetcode.com/problems/reverse-linked-list-ii/
+
+──────────────────────────────────────────────────
+
+Given the head of a singly linked list and two integers left and
+right where left <= right, reverse the nodes of the list from position
+left to position right, and return the reversed list.
+
+
+
+Example 1:
+
+Input: head = [1,2,3,4,5], left = 2, right = 4
+Output: [1,4,3,2,5]
+
+Example 2:
+
+Input: head = [5], left = 1, right = 1
+Output: [5]
+
+
+
+Constraints:
+
+        • The number of nodes in the list is n.
+
+        • 1 <= n <= 500
+
+        • -500 <= Node.val <= 500
+
+        • 1 <= left <= right <= n
+
+
+
+Follow up: Could you do it in one pass?
+"""
+
+
+# Definition for singly-linked list.
+# class ListNode:
+#     def __init__(self, val=0, next=None):
+#         self.val = val
+#         self.next = next
+class Solution:
+    def reverseBetween(
+        self, head: ListNode | None, left: int, right: int
+    ) -> ListNode | None:
+        cur, prev = head, None
+        while left > 1:
+            prev = cur
+            cur = cur.next
+            left, right = left - 1, right - 1
+
+        tail, con = cur, prev
+        while right:
+            third = cur.next
+            cur.next = prev
+            prev = cur
+            cur = third
+            right -= 1
+
+        if con:
+            con.next = prev
+        else:
+            head = prev
+
+        tail.next = cur
+        return head