LeetCode primary algorithm - list 02: Merge two ordered lists
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First, the primer
It was launched by the LeetCode official list of interview questions classics -
this module is to explore the corresponding primary algorithm - designed to help entry-algorithm. Our first pass leetcode brush is recommended topics.
View the complete algorithm Offer to prove safety issues resolved, please click on the link github:
github address
Second, the title
The two ordered lists merged into a new order list and return. The new list is by all nodes in a given mosaic composed of two lists.
Example:
输入：1->2->4, 1->3->4
输出：1->1->2->3->4->41, ideas
A number of first ordered list we see we can first compare the two lists of equal length portions, arranged in order, for the remaining part of a linked list, directly into the final list, the detailed procedure, see code.
2, programming
python
# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution(object):
def mergeTwoLists(self, l1, l2):
"""
:type l1: ListNode
:type l2: ListNode
:rtype: ListNode
"""
new_head = ListNode(0)
#返回的是合并后的列表 所以让一个节点等于这个空的节点
pHead = new_head
#进行排序
while l1 and l2:
if l1.val > l2.val:
new_head.next = l2
l2 = l2.next
else:
new_head.next = l1
l1 = l1.next
new_head = new_head.next
# 遍历剩下没遍历的列表
if l1:
new_head.next = l1
elif l2:
new_head.next = l2
return pHead.nextAIMI-CN AI learning exchange group [1015286623] for more information on AI
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