Sort a linked list using insertion sort.
Example 1:
Input: 4->2->1->3
Output: 1->2->3->4
Example 2:
Input: -1->5->3->4->0 Output: -1->0->3->4->5
思路
This question is meant to let you use the insertion sort to sort the list, according to the insertion sort an array of ideas, we can know that he is the array into two parts, sorted, unsorted. Array every time we ordered never to extract a good number will insert it into an array of sort. Until the number of unsorted array to zero. If it is applied to the list, the idea is the same and you can see a detailed illustration of steps to solve the code. The time complexity is O (n- 2 ), the spatial complexity is (1) O
illustrates the step of
Resolution code
1 # Definition for singly-linked list.
2 # class ListNode(object):
3 # def __init__(self, x):
4 # self.val = x
5 # self.next = None
6
7 class Solution(object):
8 def insertionSortList(self, head):
9 """
10 :type head: ListNode
11 :rtype: ListNode
12 """
13 if not head:
14 return head
15 cur = head
16 while cur.next: # 循环结束条件
17 if cur.next.val < cur.val: # 如果下一个节点的值比当前节点值大，则需要将其插入到相应的位置
18 tem_cur = cur.next # 将cur.next节点独立出来，
19 cur.next = cur.next.next # 后cur.next下一个节点连接起来。
20 tem_head = head
21 if tem_head.val >= tem_cur.val: # 从头节点开始查找合适的插入位置，如果cur.next的值小于头节点的值直接插入到头节点中。
22 tem_cur.next = head
23 head = tem_cur
24 else: # 否者遍历查找合适的位置
25 while tem_head.next.val < tem_cur.val:
26 tem_head = tem_head.next
27 t_next = tem_head.next # 将节点插入到合适的位置
28 tem_head.next = tem_cur
29 tem_cur.next = t_next
30 else: # 说明cur.next的值大于cur的值，更新cur的值，直接遍历下一个节点
31 cur = cur.next
32 return head # 返回头节点。