题目

Given a linked list, swap every two adjacent nodes and return its head.

For example,

Given 1->2->3->4, you should return the list as 2->1->4->3.

Your algorithm should use only constant space. You may not modify the values in the list, only nodes itself can be changed.

分析

如示例所示，给定一个链表，要求交换链表中相邻两个节点。

对于此题的程序实现，必须注意的是指针的判空，否则，一不注意就会出现空指针异常。

AC代码

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* swapPairs(ListNode* head) {
        if (head == NULL || head->next == NULL)
            return head;

        ListNode   *p = head , *q = p->next;
        //首先交换头两个结点,同时保存q后结点
        ListNode *r = q->next;

        head = q;
        p->next = r;
        q->next = p;
        if (r && r->next)
        {
            ListNode *pre = p;
            p = p->next;
            q = p->next;

            while (q)
            {
                //保存q结点后结点
                ListNode *r = q->next;

                pre->next = q;
                p->next = r;
                q->next = p;
                if (r && r->next)
                {
                    pre = p;
                    p = r;
                    q = p->next;
                }
                else{
                    break;
                }
            }
        }
        return head;
    }
};

GitHub测试程序源码