재귀 구조2 [LEET CODE] 24. Swap Nodes in Pairs # Problem Given a linked list, swap every two adjacent nodes and return its head. Constraints: The number of nodes in the list is in the range [0, 100]. 0 ListNode: cur = head while cur and cur.next: # 값만 교환 cur.val, cur.next.val = cur.next.val, cur.val cur = cur.next.next return head # Solution 2 - 반복 구조로 스왑 # Definition for singly-linked list. class ListNode: def __init__(self, x): self.val = .. 2021. 3. 23. [LEET CODE] 206. Reverse Linked List # Problem Given the head of a singly linked list, reverse the list, and return the reversed list. Constraints: The number of nodes in the list is the range [0, 5000]. -5000 ListNode: def reverse(node: ListNode, prev: ListNode = None): if not node: return prev next, node.next = node.next, prev return reverse(next, node) return reverse(head) # Solution 2 - 반복 구조로 뒤집기 # Definition for singly-linked.. 2021. 3. 22. 이전 1 다음 반응형
