边工作边刷题：70天一遍leetcode: day 9

Insertion Sort List

又是4结点互连的问题。规律就是用cur.next作为当前。

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

class Solution(object):
    def insertionSortList(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        if not head: return head
        dummy = ListNode(0)
        dummy.next = head
        cur = head
        while cur.next:
            if cur.next.val<cur.val:
                pre = dummy
                while pre!=cur and pre.next.val<cur.next.val:
                    pre=pre.next
                next = pre.next
                pre.next = cur.next
                cur.next = cur.next.next
                pre.next.next = next
            else:
                cur=cur.next
        return dummy.next