题目:原题链接(困难)
标签:链表、链表-双指针
解法 | 时间复杂度 | 空间复杂度 | 执行用时 |
---|---|---|---|
Ans 1 (Python) | 56ms (69.08%) | ||
Ans 2 (Python) | |||
Ans 3 (Python) |
LeetCode的Python执行用时随缘,只要时间复杂度没有明显差异,执行用时一般都在同一个量级,仅作参考意义。
解法一(双指针):
def reverseKGroup(self, head: ListNode, k: int) -> ListNode:
ans = node = ListNode(0)
ans.next = head
while node:
# 判断链表剩余长度是否充足
curr = node
is_enough = True
for _ in range(k):
if curr is None or curr.next is None:
is_enough = False
break
curr = curr.next
if not is_enough:
break
# 翻转链表
curr = node.next
for _ in range(k - 1):
now = ListNode(curr.next.val)
now.next = node.next
node.next = now
curr.next = curr.next.next
node = curr
return ans.next