【c++】代码随想录算法训练营day4| 24. 两两交换链表中的节点 ● 19.删除链表的倒数第N个节点 【附注释：踩的坑】

24. 两两交换链表中的节点

 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* swapPairs(ListNode* head) {
        ListNode* dummy = new ListNode(0);
        dummy->next = head;
        ListNode* cur = dummy;
        while(cur->next != NULL && cur->next->next != NULL){    //先后顺序不能乱，否则会报空指针异常。
            ListNode* tmp = cur->next;
            ListNode* tmp1 = cur->next->next->next; //加入交换节点1和2，节点1和3会容易丢失，先保存。
            cur->next = cur->next->next;
            cur->next->next = tmp;
            tmp->next = tmp1;
            cur = cur->next->next;  //cur要向后移动两位，继续对后面的节点进行操作
        }
    return dummy->next;
    }
};

19.删除链表的倒数第N个节点

 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        ListNode* dummy = new ListNode(0);
        dummy->next = head;
        ListNode* slow = dummy;
        ListNode* fast = dummy;
        while(n-- && fast != NULL){
            fast = fast->next;
        }
        fast = fast->next;  //快指针必须再先走一步，要让慢指针指向删除的那个节点的上一个
        while(fast != NULL){
            fast = fast->next;
            slow = slow->next;
        }
        ListNode* tmp =slow->next;
        slow->next = slow->next->next;
        delete tmp;  //还要把这个节点从内存中删掉
        return dummy->next;
        
    }
};