链表_leetcode148-链表归并排序

# Definition for singly-linked list.
class ListNode:
def __init__(self, x):
self.val = x
self.next = None


class Solution:
def sortList(self, head):
"""
        :type head: ListNode
        :rtype: ListNode
        """
        if not head or not head.next:
return head

        mid = self.getMid(head)

        l = head
        r = mid.next
        mid.next =None
        return self.merge(self.sortList(l),self.sortList(r))

def getMid(self,head):
if not head :
return head
        fast = slow = head

while fast.next and fast.next.next:
            slow = fast.next
            fast = fast.next.next

return slow


def merge(self,p,q):
        dummyHead = ListNode(0)
        pre = dummyHead
while p and q:
if p.val < q.val:
                pre.next = p
                pre = pre.next
                p = p .next
else:
                pre.next = q
                pre = pre.next
                q = q.next

if p:
            pre.next = p
else:
            pre.next = q

return dummyHead.next

转载于:https://www.cnblogs.com/lux-ace/p/10557221.html