Swap Nodes in Pairs

时间：2019-08-27 00:10:29 收藏：0 阅读：101

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

Example:

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

Note:

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

class Solution{
    public:
        ListNode* swapPairs(ListNode* head){
            ListNode *dummy = new ListNode(-1);
            ListNode *pre = dummy;
            
            dummy->next = head;
            while(pre->next && pre->next->next){
                ListNode *t = pre->next->next;
                pre->next->next = t->next;
                t->next = pre->next;
                pre->next = t;
                pre = t->next;  
            }
            return dummy->next;
        }
};

//递归写法稍微有些复杂
class Solution{
    public:
        ListNode* swapPairs(ListNode* head){
            if(!head || !head->next) return head;
            ListNode *t = head->next;
            head->next = swapPairs(head->next->next);
            t->next = head;
            return t;
        }
};

原文：https://www.cnblogs.com/hujianglang/p/11415658.html