[LinkedList]021 Merge Two Sorted Lists

• 分类：LinkedList

• 考察知识点：LinkedList

• 最优解时间复杂度：**O(n) **

21. Merge Two Sorted Lists

Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.

Example:

Input: 1->2->4, 1->3->4
Output: 1->1->2->3->4->4

代码：

我的解法:

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

class Solution:
    def mergeTwoLists(self, l1, l2):
        """
        :type l1: ListNode
        :type l2: ListNode
        :rtype: ListNode
        """
        res=ListNode(0)
        p1=l1
        p2=l2
        p=res
        while(p1!=None and p2!=None):
            if p1.val<=p2.val:
                p.next=p1
                p1=p1.next
                p=p.next
            else:
                p.next=p2
                p2=p2.next
                p=p.next
                    
        if p1==None:
                p.next=p2
        elif p2==None:
                p.next=p1
                
        return res.next

Beautiful解法:

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

class Solution:
    def mergeTwoLists(self, l1, l2):
        """
        :type l1: ListNode
        :type l2: ListNode
        :rtype: ListNode
        """
        res=ListNode(0)
        p1=l1
        p2=l2
        p=res
        while(p1 and p2):
            if p1.val<=p2.val:
                p.next=p1
                p1=p1.next         
            else:
                p.next=p2
                p2=p2.next
            p=p.next
                    

        p.next=p2 or p1

                
        return res.next

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讨论：

1.这道题我的解法是对的(很好，虽然是道easy题)
2.但是我的代码不是很优雅不是很beautiful!把代码改的beautiful后性能提升了很多！

我的版本

beautiful版本