LeetCode Hot100 NO.24

Problem

https://leetcode.cn/problems/swap-nodes-in-pairs/description/?envType=study-plan-v2&envId=top-100-liked

Solution

两两交换链表中的节点
pre->p1->p2->p3 => pre->p2->p1->p3
设置一个dummy节点避免对头节点的特殊处理，遍历整个链表完成上面的操作，遍历过程要保证p1p2非空
可以只记录p2

/**
 * 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(0, head);

        ListNode *pre = &dummy;

        while (pre->next && pre->next->next)
        {
            ListNode *p2 = pre->next->next;
            pre->next->next = p2->next;
            p2->next = pre->next;
            pre->next = p2;

            pre = pre->next->next;
        }

        return dummy.next;
    }
};