首页 > 代码库 > 24. Swap Nodes in Pairs
24. Swap Nodes in Pairs
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.
1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * ListNode *next; 6 * ListNode(int x) : val(x), next(NULL) {} 7 * }; 8 */ 9 class Solution {10 public:11 ListNode* swapPairs(ListNode* head) {12 13 if(head == NULL || head->next == NULL){14 return head;15 }16 ListNode* newHead = new ListNode(-1);17 ListNode* tail = newHead;18 newHead->next = head;19 20 while(tail && tail->next && tail->next->next){21 ListNode* n = tail->next;22 ListNode* nn = tail->next->next;23 24 //tail->next = nn;25 n->next = nn->next;26 nn->next = n;27 tail->next = nn;28 29 tail = tail->next->next;30 }31 32 return newHead->next;33 }34 };
24. Swap Nodes in Pairs
声明：以上内容来自用户投稿及互联网公开渠道收集整理发布，本网站不拥有所有权，未作人工编辑处理，也不承担相关法律责任，若内容有误或涉及侵权可进行投诉： 投诉/举报 工作人员会在5个工作日内联系你，一经查实，本站将立刻删除涉嫌侵权内容。