24. 两辆交换链表中的节点

题目

卡哥的讲解很详细了

卡哥视频讲解一如既往的把小细节都讲到了

跟着卡哥的代码敲了下

/**
 * 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 *dummyHead = new ListNode(0);
        dummyHead->next = head;
        ListNode *cur = dummyHead;
        while (cur->next != nullptr && cur->next->next != nullptr)
        {
            ListNode *tmp = cur->next;
            ListNode *tmp1 = cur->next->next->next;

            cur->next = cur->next->next;
            cur->next->next = tmp;
            cur->next->next->next = tmp1;

            cur = cur->next->next;
        }
        ListNode *result = dummyHead->next;
        delete dummyHead;
        return result;
    }
};

再附上自己刚开始理解过程中的一个误区