题目分析:
这一题是把有序链表每相邻的k个值倒序,最后不足k个值就直接保持输出。这一题比较有难度,单独写一个反序函数会比较明了
代码说明:
1、reverse
是反序函数,如k=4时它的操作过程如下(输入为1 -> 2 -> 3 -> 4 -> 5,0是它附加的头):
0 -> 2 -> 1 -> 3 -> 4 -> 5
0 -> 3 -> 2 -> 1 -> 4 -> 5
0 -> 4 -> 3 -> 2 -> 1 -> 5
倒序的原理就是1234 -> 2134 -> 3214 -> 4321这样变化的。其中start,end,newhead
三个指针指向都是不变的,只是每次将start后面的元素移动到newhead后面,直到newhead后面是end就倒序完成了。
2、每次选取k个元素进入倒序函数,如果剩余不够k个,可以直接返回头部。
for i in range(k-1):
end=end.next
if end.next==None: return nhead.next
res=self.reverse(start.next, end.next)
3、令开始指针下一个节点指向返回回来的倒序链表,再将开始指针指向返回倒序链表的尾部准备下一次循环。
start.next=res[0] start=res[1]
测试代码:
def print_ListNode(head):
while (True):
print(head.val, end='')
if head.next is not None:
head = head.next
else:
break
print(' -> ', end='')
print()
class ListNode:
def __init__(self, x):
self.val = x
self.next = None
list_node1_1 = ListNode(1)
list_node1_2 = ListNode(2)
list_node1_3 = ListNode(3)
list_node1_4 = ListNode(4)
list_node1_5 = ListNode(5)
list_node1_1.next = list_node1_2
list_node1_2.next = list_node1_3
list_node1_3.next = list_node1_4
list_node1_4.next = list_node1_5
class Solution:
def reverse(self, start, end):
a = start.val
b = end.val
newhead=ListNode(0); newhead.next=start
while newhead.next!=end:
tmp=start.next
start.next=tmp.next
tmp.next=newhead.next
newhead.next=tmp
return [end, start]
def reverseKGroup(self, head, k):
if head==None: return None
nhead=ListNode(0); nhead.next=head; start=nhead
while start.next:
end=start
for i in range(k-1):
end=end.next
if end.next==None: return nhead.next
res=self.reverse(start.next, end.next)
start.next=res[0]
start=res[1]
return nhead.next
print_ListNode(Solution().reverseKGroup(list_node1_1, 3))
提交代码:
class Solution:
def reverse(self, start, end):
a = start.val
b = end.val
newhead=ListNode(0); newhead.next=start
while newhead.next!=end:
tmp=start.next
start.next=tmp.next
tmp.next=newhead.next
newhead.next=tmp
return [end, start]
def reverseKGroup(self, head, k):
if head==None: return None
nhead=ListNode(0); nhead.next=head; start=nhead
while start.next:
end=start
for i in range(k-1):
end=end.next
if end.next==None: return nhead.next
res=self.reverse(start.next, end.next)
start.next=res[0]
start=res[1]
return nhead.next