24. Swap Nodes in Pairs

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.

 

思路：

　　以1->2->3->4为例，为了方便计算首先添加一个头节点0，此时链表为0->1->2->3->4。题目要求两两节点相互交换，则最终的链表应该为0->2->1->4->3。

　　首先关注节点1和2的交换。节点0为prev，节点1为cur，具体操作步骤为：

• pre->next = cur->next;
• cur->next = cur->next->next;
• pre->next->next = cur;

　　通过这样的操作后，此时链表为0->2->1->3->4。接下来是节点3和节点4的交换操作。具体步骤与节点1和节点2的交换步骤相一致，为了重复这一过程，需要更新prev和cur。具体操作步骤为：

• pre = cur;
• cur = cur->next;

 

代码：

class Solution {
public:
    ListNode* swapPairs(ListNode* head) {
        if (head == NULL || head->next == NULL)
            return head;
        ListNode* newhead = new ListNode(0);
        newhead->next = head;
        ListNode* pre = newhead;
        ListNode* cur = head;
        while (cur && cur->next) {
            pre->next = cur->next;
            cur->next = cur->next->next;
            pre->next->next = cur;
            pre = cur;
            cur = cur->next;
        }
        return newhead->next;
    }
};