给你一个链表,两两交换其中相邻的节点,并返回交换后链表的头节点。你必须在不修改节点内部的值的情况下完成本题(即,只能进行节点交换)。
示例 1:

**输入:**head = [1,2,3,4]
输出:[2,1,4,3]
示例 2:
**输入:**head = []
输出:[]
示例 3:
**输入:**head = [1]
输出:[1]
提示:
- 链表中节点的数目在范围
[0, 100] 内
0 <= Node.val <= 100
![[Pasted image 20250405120249.png]]
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class Solution {
public:
ListNode* swapPairs(ListNode* head) {
ListNode* dummy_node = new ListNode(0);
dummy_node->next = head;
ListNode *cur = dummy_node;
while(cur ->next != nullptr && cur ->next ->next != nullptr)
{
ListNode* tmp = cur->next;
ListNode* tmp1 = cur ->next->next->next;
cur ->next = cur->next->next;
cur->next->next = tmp;
cur->next->next->next = tmp1;
cur = cur->next->next;
}
ListNode* result = dummy_node ->next;
delete dummy_node;
return result;
}
};
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