题目讲解

24. 两两交换链表中的节点

算法讲解

只需要模拟这个过程就行了，但是需要注意空指针的问题，特别是nnext指针

/**
 * 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) {
        if(head == nullptr || head->next == nullptr)return head;
        ListNode* newhead = new ListNode(-1);
        newhead->next = head;
        ListNode* prev = newhead;
        ListNode* cur = head;
        ListNode* Next = head->next;
        ListNode* nnext = Next->next;
        while(cur && Next)
        {
            prev->next = Next;
            Next->next = cur;
            cur->next = nnext;
            prev = cur;
            cur = nnext;
            if(cur)Next = cur->next;
            if(Next)nnext = Next->next;
        }
        prev = newhead->next;
        delete newhead;
        return prev;
    }
};