Description

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.

Code

1. 原始版 72ms

class Solution(object):
    def mergeTwoLists(self, l1, l2):
        head = ListNode(0)
        tmp = head
        while l1 != None
 
 and l2 != None:
            if l1.val < l2.val:
                tmp.next = ListNode(l1.val)
                l1 = l1.next
            else:
                tmp.next = ListNode(l2.val)
                l2 = l2.next
            tmp = tmp.next
        if l1 != None:
            l = l1
        else:
            l = l2
        while
 
 l != None:
            tmp.next = ListNode(l.val)
            tmp = tmp.next
            l = l.next
        return head.next

2.  優化版 52ms

class Solution(object):
    def mergeTwoLists(self, l1, l2):
        head = ListNode(0)
        tmp = head
        while l1 != None and l2 != None:
            if
 
 l1.val < l2.val:
                tmp.next = l1
                l1 = l1.next
            else:
                tmp.next = l2
                l2 = l2.next
            tmp = tmp.next
        if l1 != None:
            tmp.next = l1
        else:
            tmp.next = l2
        return head.next

Note

1. 思路
       其實根本就不用複製出一條新的連結串列存放結果嘛，直接在原來的兩條鏈上改造就是了（第6~11行）。
       對於某一條鏈中剩下的部分，不必一一複製，將現有結果的最後一個結點的next指向那條鏈中剩下的部分的頭節點即可（第13~16行）