题解 | 两两交换链表中的结点

/**
 * struct ListNode {
 *  int val;
 *  struct ListNode *next;
 *  ListNode(int x) : val(x), next(nullptr) {}
 * };
 */
class Solution {
  public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     *
     * @param head ListNode类
     * @return ListNode类
     */
    ListNode* swapPairs(ListNode* head) {
        ListNode hair(0), *cur = &hair;
        hair.next = head;

        while (cur->next && cur->next->next) {
            ListNode* cur1 = cur->next;
            ListNode* cur2 = cur1->next;

            cur->next = cur2;
            cur1->next = cur2->next;
            cur2->next = cur1;
            cur = cur1;
        }

        return hair.next;
    }
};