题解 | 两两交换链表的节点

/**
 * 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* swapLinkedPair(ListNode* head) {
        if(!head || !head->next)return head;

        ListNode dummy(0);
        dummy.next = head;  

        ListNode *node = dummy.next->next;
        ListNode *currhead = &dummy;

        while(node){   
            //jiaohuan
            currhead->next->next = node->next;
            node->next = currhead->next;
            currhead->next = node;   

            currhead = currhead->next->next;   //更新下一组的头节点；
            node = currhead;					//更新用于检测节点，也是交换节点；

            if(node->next){  //在运行时检测是否有两个成一组
                node = node->next;
            }else{
                break;
            }
            node = node->next;
        }
        return dummy.next;
    }
};