Question

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

a = ListNode(5)
b = ListNode(10)
c = ListNode(20)


e = ListNode(0)
f = ListNode(5)
g = ListNode(21)
h = ListNode(30)


a.next = b
b.next = c

e.next = f
f.next = g
g.next = h

因此，我有两个头为a和e的单链接列表

我想按其值的升序合并。现在，我要合并它们，以遍历两个链表，比较值，直到其中一个链表到达None（其中一个链表比另一个链表短，所以一个将到达{{1} }之前）

None

因此，这是我理想中要发生的事情，但显然不会发生，正如您在打印报表时所看到的！

class Solution: def mergeTwoLists(self, l1, l2): tempNode = ListNode(0) returnNode = tempNode while (l1.next != None) and (l2.next != None): if l1.val < l2.val: print("l1.val < l2.val", l1.val, l2.val) tempNode.next = l1 tempNode = tempNode.next l1 = l1.next elif l1.val == l2.val: print("l1.val == l2.val", l1.val, l2.val) #If both the values are equal, assign l1's value first #then make l2's value follow l1's value using tempNode tempNode.next = l1 tempNode = tempNode.next #Because of the previous statement, after execution of this statement, I'm assuming tempNode.val is now l1.val and tempNode.next = l1.next #tempNode.next is supposed to be equal to l1.next, instead assign it to l2 tempNode.next = l2 tempNode = tempNode.next #Because of the previous statement, after execution of this statement, I'm assuming tempNode.val is now l2.val and tempNode.next = l2.next #Increment both l1 and l2 l1 = l1.next l2 = l2.next else: print("l1.val > l2.val", l1.val, l2.val) tempNode.next = l2 tempNode = tempNode.next l2 = l2.next sol = Solution() sol.mergeTwoLists(a, e)

增加或前进l2！

l1.val = 5 > l2.val = 0，即l1.val = 5 ==

它们都相等，因此将l2.val == 5移至下一个，并将l1移至下一个

现在，l2应该是l1.val，而10应该是l2.val，但是 21被打印为l1.val，而5被打印为l2.val并陷入无限循环。

为什么21停留在l1.val而不是更新为5，为什么它停留在这个无限循环中而不是其中一个依我的{{1 }}语句？

Answer 1

问题在下面的代码片段中

tempNode.next = l1 
tempNode = tempNode.next 
tempNode.next = l2
tempNode = tempNode.next
l1 = l1.next
l2 = l2.next

只需将其更改为以下内容，您的代码即可使用

tempNode.next = l1 
tempNode = tempNode.next
l1 = l1.next 
tempNode.next = l2
tempNode = tempNode.next
l2 = l2.next

您的方法存在的问题是，当您执行tempNode.next = l2时，您正在修改ListNode所指向的实际l1，这会使您陷入无限循环之中

Answer 2

您需要将l1和l2的值分配给tempNode.val，而不是将l1和l2节点本身分配给tempNode'的下一个节点。如果其他列表为空，则还需要将l1或l2的剩余值添加到tempNode中。

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

a = ListNode(5)
b = ListNode(10)
c = ListNode(20)


e = ListNode(0)
f = ListNode(5)
g = ListNode(21)
h = ListNode(30)


a.next = b
b.next = c

e.next = f
f.next = g
g.next = h

class Solution:
    def mergeTwoLists(self, l1, l2):

        returnNode = tempNode = ListNode()

        while l1 or l2:
            if not l1:
                print('l1 is empty; adding value from l2:', l2.val)
                tempNode.val = l2.val
                tempNode.next = ListNode()
                tempNode = tempNode.next
                l2 = l2.next
            elif not l2:
                print('l2 is empty; adding value from l1:', l1.val)
                tempNode.val = l1.val
                tempNode.next = ListNode()
                tempNode = tempNode.next
                l1 = l1.next
            elif l1.val < l2.val:

                print("l1.val < l2.val", l1.val, l2.val)
                tempNode.val = l1.val

                tempNode.next = ListNode()
                tempNode = tempNode.next
                l1 = l1.next

            elif l1.val == l2.val:

                print("l1.val == l2.val", l1.val, l2.val)

                #If both the values are equal, assign l1's value first
                #then make l2's value follow l1's value using tempNode

                tempNode.val = l1.val
                tempNode.next = ListNode() #Because of the previous statement, after execution of this statement, I'm assuming tempNode.val is now l1.val and tempNode.next = l1.next
                tempNode = tempNode.next

                #tempNode.next is supposed to be equal to l1.next, instead assign it to l2
                tempNode.val = l2.val
                tempNode.next = ListNode() #Because of the previous statement, after execution of this statement, I'm assuming tempNode.val is now l2.val and tempNode.next = l2.next
                tempNode = tempNode.next

                #Increment both l1 and l2
                l1 = l1.next
                l2 = l2.next


            else:
                print("l1.val > l2.val", l1.val, l2.val)
                tempNode.val = l2.val

                tempNode.next = ListNode()
                tempNode = tempNode.next
                l2 = l2.next
        return returnNode

sol = Solution()
node = sol.mergeTwoLists(a, e)
while node and node.val is not None:
    print(node.val)
    node = node.next

这将输出：

l1.val > l2.val 5 0
l1.val == l2.val 5 5
l1.val < l2.val 10 21
l1.val < l2.val 20 21
l1 is empty; adding value from l2: 21
l1 is empty; adding value from l2: 30
0
5
5
10
20
21
30

在Python中以升序合并两个排序的链表：单链表指针更新问题

2 个答案: