leetcode 024

 

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.

 

             head

dimmy -> 1 --> 2 --> 3 --> 4  对cur先后再前

  pre       cur  

 

class Solution {
public:
    ListNode* swapPairs(ListNode* head) {
        if(head == NULL)
            return head;
        ListNode *dummy = new ListNode(0);
        dummy->next = head;
        ListNode *pre = dummy, *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 dummy->next;
        
    }
};

  

posted @ 2016-07-07 20:21  路萧  阅读(172)  评论(0编辑  收藏  举报