cur_a, cur_b = headA, headB
while cur_a != cur_b:
cur_a = cur_a.next if cur_a else headB
cur_b = cur_b.next if cur_b else headA
return cur_a
# 此外，还有一种方法
# https://discuss.leetcode.com/topic/26018/python-solution-o-n-time-and-o-1-space
# 这种方法相比上两种稍微难理解点，因为最后的结果的正确性需要证明下，代码也更多一些，而且也没有运行优势
# 因此不再列出
if __name__ == '__main__':
print _solve1(None, None)
print _solve1(ListNode(1), None)
print _solve1(None, ListNode(1))
print _solve1(ListNode(1), ListNode(1))
HEAD1 = ListNode.from_list([4, 5, 6, 7, 8, 0, 1])
HEAD2 = ListNode.from_list([1, 2, 3])
HEAD2.next = HEAD1.next.next.next
HEAD1.print_list()
HEAD2.print_list()
print _solve1(HEAD1, HEAD2)
print _solve1(HEAD2, HEAD1)
HEAD2 = ListNode.from_list([1, 2, 3])
HEAD2.next.next = HEAD1.next.next
HEAD1.print_list()
HEAD2.print_list()
print _solve1(HEAD1, HEAD2)
print _solve1(HEAD2, HEAD1)
l1 = l1.next
else:
cur.next = l2
l2 = l2.next
cur = cur.next
cur.next = l1 if l1 else l2
return dummy.next
def _solve1(l1, l2):
def _help(l1, l2):
if not l1:
return l2
if not l2:
return l1
if l1.val < l2.val:
l1.next = _help(l1.next, l2)
return l1
else:
l2.next = _help(l1, l2.next)
return l2
return _help(l1, l2)
if __name__ == '__main__':
l1 = ListNode.from_list([1, 3])
l2 = ListNode.from_list([0, 4])
_solve(l2, l1).print_list()
# coding=utf-8
"""Remove Linked List Elements."""
from ListNode import ListNode
def _solve(head, val):
dump = ListNode('a')
dump.next = head
prev, cur = dump, head
while cur:
if cur.val == val:
prev.next = cur.next
cur = cur.next
else:
prev, cur = cur, cur.next
return dump.next
if __name__ == '__main__':
print _solve(None, 0)
_solve(ListNode(1), 0).print_list()
print _solve(ListNode(1), 1)
print _solve(ListNode.from_list([1, 1, 1, 1]), 1)
_solve(ListNode.from_list([1, 2, 6, 4, 5, 6, 6]), 6).print_list()
if not head:
return True
length = 0
cur = head
while cur:
cur = cur.next
length += 1
prev, next = None, None
cur = head
for _ in xrange(0, length / 2):
next = cur.next
cur.next = prev
prev, cur = cur, next
if length & 1 == 1:
cur = cur.next
while cur:
if cur.val != prev.val:
return False
cur = cur.next
prev = prev.next
return True
if __name__ == '__main__':
print _solve(None)
print _solve(ListNode(1))
print _solve(ListNode.from_list([1, 2, 1]))
print _solve(ListNode.from_list([1, 2, 2]))
print _solve(ListNode.from_list([1, 2, 2, 1]))
print _solve(ListNode.from_list([1, 2, 2, 2]))
slow = slow.next
quick = quick.next and quick.next.next
return quick == slow
# 翻转列表，如果有环，翻转列表后，一定会回到头部
# 这种方法会破坏原 NodeList
def _solve1(head):
prev, cur, next = None, head, None
while cur and cur.next:
if cur.next == head:
return True
next = cur.next
cur.next = prev
prev, cur = cur, next
return False
if __name__ == '__main__':
RECYCLE_NODE = ListNode.from_list([1, 2, 3])
RECYCLE_NODE.next.next.next = RECYCLE_NODE
node1 = ListNode(1)
node1.next = node1
print _solve1(RECYCLE_NODE)
print _solve(node1)
print _solve1(ListNode.from_list([1, 2, 3]))
print _solve1(ListNode(1))
print _solve1(None)
slow = fast = head
while fast.next and fast.next.next:
slow = slow.next
fast = fast.next.next
head1, head2 = head, slow.next
slow.next = None
head1 = _sort(head1)
head2 = _sort(head2)
prev = faker
while head1 and head2:
if head1.val < head2.val:
prev.next = head1
head1 = head1.next
else:
prev.next = head2
head2 = head2.next
prev = prev.next
if head1:
prev.next = head1
else:
prev.next = head2
return faker.next
return _sort(head)
if __name__ == '__main__':
_solve(ListNode.from_list(range(1, 5001)))
_solve1(ListNode.from_list(range(1, 5001)))
