leetcode 刷题之路 87 Swap Nodes in Pairs

BurningWall

于 2014-08-13 16:17:59 发布

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本文介绍了一种算法，用于两两交换链表中的相邻节点，仅使用常数空间复杂度，不修改链表值而改变节点指向实现该功能。

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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.

两两交换链表中的相邻节点。

accepted solution：

/**
 * 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 *dummyHead=new ListNode(0);
        dummyHead->next=head;
        ListNode *previous=dummyHead,*current=previous->next;
        while(current!=NULL&¤t->next!=NULL)
        {
            previous->next=current->next;
            current->next=previous->next->next;
            previous->next->next=current;
            previous=current;
            current=current->next;
        }
        current=dummyHead->next;
        delete dummyHead;
        return current;
    }
};