代码随想录算法训练营第四天|24. 两两交换链表中的节点

链表没有学好，写了一小时才写出来

 

有几个心得：

1. 不存在指针被复制之后就是地址完全被拷贝了，考虑的有点多

2. current.next = right这句话完全忘记加了，debug的主要时间都在这，导致一直出的[1,3]，没有考虑交换之后，前一个结点的指向也要改变。

 

 

 

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def swapPairs(self, head: Optional[ListNode]) -> Optional[ListNode]:
        dummyhead = ListNode(next = head)
        current = dummyhead

        if current.next == None:
            return current.next
        

        while current.next != None:
            if current.next.next:
                left = current.next
                right = current.next.next
                if current == dummyhead:
                    dummyhead.next = current.next
                print("left.val,",left.val)
                print("right.val,",right.val)
                print("current.val,",current.val)
                left.next = right.next
                # print("current.val,",current.val)
                right.next = current.next

                # print("current.val,",current.val)
                # print("current.next.val,",current.next.val)
                current.next = right
                current = current.next.next
                print("left.val,",left.val)
                print("right.val,",right.val)
                print("current.val,",current.val)
                print("dummyhead.next.val,",dummyhead.next.val)
            else:
                break
        return dummyhead.next