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.
这道题不算难,是基本的链表操作题,我们可以分别用递归和迭代来实现。对于迭代实现,还是需要建立dummy节点,注意在连接节点的时候,最好画个图,以免把自己搞晕了,参见代码如下:
解法一:
class Solution { public: ListNode* swapPairs(ListNode* head) { ListNode *dummy = new ListNode(-1), *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; } };
递归的写法就更简洁了,实际上利用了回溯的思想,递归遍历到链表末尾,然后先交换末尾两个,然后依次往前交换:
解法二:
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; } };
本文转自博客园Grandyang的博客,原文链接:成对交换节点[LeetCode] Swap Nodes in Pairs ,如需转载请自行联系原博主。