36. 翻转链表 II
翻转链表中第m个节点到第n个节点的部分
样例
给出链表1->2->3->4->5->null， m = 2 和n = 4，返回1->4->3->2->5->null
挑战
在原地一次翻转完成
注意事项
m，n满足1 ≤ m ≤ n ≤ 链表长度
实现代码：
"""
Definition of ListNode
class ListNode(object):
def __init__(self, val, next=None):
self.val = val
self.next = next
"""
class Solution:
"""
@param head: ListNode head is the head of the linked list
@param m: An integer
@param n: An integer
@return: The head of the reversed ListNode
"""
'''
def reverseBetween(self, head, m, n):
# write your code here
#将链表转成数组，取巧的方法
cur = head
temp_list = []
while cur:
temp_list.append(cur.val)
cur = cur.next
changed_list = temp_list[m-1:n]
changed_list.reverse()
reversed_list = temp_list[:m-1] + changed_list + temp_list[n:]
dummy = ListNode(0)
pre = dummy
for item in reversed_list:
pre.next = ListNode(item)
pre = pre.next
return dummy.next
'''
def reverseBetween(self, head, m, n):
# write your code here
if m == n:
return head
dummy = ListNode(0)
dummy.next = head
#翻转部分
mpre = dummy
mcur = head
#定位到翻转部分
for i in range(m - 1):
mpre = mpre.next
mcur = mpre.next
#定义进行交换的拷贝变量
temp_cur = mcur
temp_post = temp_cur.next
#对n-m的结点进行翻转
for i in range(n-m):
tmp = temp_post.next
temp_post.next = temp_cur
temp_cur = temp_post
temp_post = tmp
#更新交换后的结点
mpre.next = temp_cur
mcur.next = temp_post
return dummy.next