[LeetCode] Swap Nodes in Pairs

Swap Nodes in Pairs

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.

解题思路：

注意头结点的不同。建议先创造一个伪头结点，其next指向head，这样，之后所有的操作都是一样的了。这样能够简化代码。

/**
 * 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) {
        ListNode* myHead=new ListNode(0);
        myHead->next=head;
        ListNode* pre=myHead;
        while(pre->next!=NULL&&pre->next->next!=NULL){
            ListNode* p=pre->next;
            pre->next=p->next;
            p->next=pre->next->next;
            pre->next->next=p;
            
            pre=pre->next->next;
        }
        head=myHead->next;
        delete myHead;
        return head;
    }
};

[LeetCode] Swap Nodes in Pairs

原文：http://blog.csdn.net/kangrydotnet/article/details/45220539