题解 | #链表内指定区间反转#
链表内指定区间反转
https://www.nowcoder.com/practice/b58434e200a648c589ca2063f1faf58c
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
#
# 代码中的类名、方法名、参数名已经指定,请勿修改,直接返回方法规定的值即可
#
#
# @param head ListNode类
# @param m int整型
# @param n int整型
# @return ListNode类
#
class Solution:
def reverseBetween(self , head: ListNode, m: int, n: int) -> ListNode:
# write code here
if head==None or head.next==None:return head
late = n-m
pi = head
for _ in range(late):pi = pi.next
pj = head
dummy_head = ListNode(0)
dummy_head.next = head
prej = dummy_head
for _ in range(m-1):
pi = pi.next
pj = pj.next
prej = prej.next
#反转
nexti = pi.next
pre,p = pj,pj.next
while(p!=nexti):
tmp = p
p = p.next
tmp.next = pre
pre = tmp
prej.next = pi
pj.next = nexti
return dummy_head.next
指定区间翻转,本来以为挺简单就能搞定,但是其实发现有很多细节的地方没注意就会报错,比如边界条件的处理,然后要做一个假头,后面的指针比前面的多走多少步,后面又应该一起走多少步,然后最后翻转的这个操作怎么写才不会错,然后就是debug要快……
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