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 # ''' 通过将链表转换到列表中实现切片翻转, 然后再讲结果放到链表中,返回第一个节点''' class Solution: def reverseBetween(self , head: ListNode, m: int, n: int) -> ListNode: data = [] if head ==None: return head while head: data.append(head.val) head = head.next df = data[0:m-1]+data[m-1:n][::-1]+data[n:] res = ListNode(df[0]) pre = res for i in range(1, len(df)): res.next = ListNode(df[i]) res = res.next return pre
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