Leetcode 24

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Question

Solution

根据题意，不改变结点的值，改变结点的位置。

创建一个指针，需要判断其之后的以及之后的之后的结点是否存在。如果存在，调整顺序.

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

class Solution:
    def swapPairs(self, head: ListNode) -> ListNode:
        ans = cur = ListNode(0)
        ans.next = head
        while cur.next and cur.next.next:
            a = cur.next
            b = a.next
            a.next, b.next, cur.next = b.next, a, b
            cur = cur.next.next
            
        return ans.next