题解 | 两两交换链表中的结点

迭代法

/**
 * 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* swapPairs(ListNode* head) {
        // write code here
        ListNode* dummy = new ListNode(0);
        dummy->next = head;
        ListNode* prev = dummy;
        while(prev->next != nullptr && prev->next->next !=nullptr)
        {
            ListNode* first = prev->next;
            ListNode* second = first->next;
            first->next = second->next;
            second->next = first;
            prev->next = second;
            prev = first;
        }
        return dummy->next;
    }
};