总结
- 有序链表的合并【python版】
- LawsonAbs
- LeetCode 链接
思路：
- step1. 先比较两条链表谁的头结点对应的值最大，取较小的作为待返回的头结点
- step2. 依次比较两条链表，然后将小的值对应的结点放到结果链表末尾
- step3. 重复step2直到某条链表完成比较
- step4. 将未比较过的值直接链接到结果链表中
# Definition for singly-linked list.
class ListNode:
def __init__(self, val=0, next=None):
self.val = val
self.next = next
class Solution:
def mergeTwoLists(self, list1, list2):
h1 = list1
h2 = list2
if h1 is None :
return h2
if h2 is None:
return h1
# step1. 先确定头是谁
start = h_res = None
if h1.val <= h2.val :
start = h1
h1 = h1.next
else:
start = h2
h2 = h2.next
h_res = start
# step2. 再依次遍历，合并两条链表
while(h1 is not None and h2 is not None):
if h1.val <= h2.val:
tmp = h1.next # 保存
h_res.next = h1
h1 = tmp
else:
tmp = h2.next
h_res.next = h2
h2 = tmp
h_res = h_res.next
if h1 is not None:
h_res.next = h1
if h2 is not None:
h_res.next = h2
return start
node3 = ListNode(val=3,next = None)
node2 = ListNode(val=2,next = node3)
node1 = ListNode(val=1,next = node2)
node6 = ListNode(val=6,next = None)
node5 = ListNode(val=5,next = node6)
node4 = ListNode(val=2,next = node5)
s = Solution()
s.mergeTwoLists(node1,node4)
