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.
本文只需要两个指针依次交换相互位置。但是需要注意单数链表的情况。时间：4ms
代码如下：
/** * 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) { if (head == NULL || head->next == NULL) return head; ListNode *p, *q, *phead, *pnext=new ListNode(0); p = head; q = head->next; phead = pnext; do{ q = p->next; p->next = q->next; q->next = p; pnext->next = q; pnext = p; if (p != NULL) p = p->next; } while (p != NULL&&p->next != NULL); return phead->next; } };
转载于:https://www.cnblogs.com/Scorpio989/p/4547807.html
