Given a linked list, swap every two adjacent nodes and return its head.
For example,
Given 1->2->3->4, you should return the list as 2->1->4->3.
Your algorithm should use only constant space. You may not modify the values in the list, only nodes itself can be changed.
用三个指针p, q, pre, p指向当前节点，初始化为head，q指向p的下一个节点，pre指向上次完成的节点，初始化null
交换p，q，同时需要修改pre的下一个节点(之前指向p，由于p和q交换位置，需要修改指向q):
while (p && p->next) {
ListNode *q = p->next;
p->next = q->next;
q->next = p;
if (pre)
pre->next = q;
pre = p;
p = p->next;
}同时需要注意更新head指针。完整代码:
ListNode *swapPairs(ListNode *head) {
if (head == nullptr || head->next == nullptr)
return head;
ListNode *p = head;
head = p->next;
ListNode *pre = nullptr;
while (p && p->next) {
ListNode *q = p->next;
p->next = q->next;
q->next = p;
if (pre)
pre->next = q;
pre = p;
p = p->next;
}
return head;
}