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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.
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/** * 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* help = new ListNode(0); help->next = head; ListNode *first = help, *second = head, *third = head->next; while (third != NULL) { ListNode* tmp = third->next; third->next = first->next; // 调整一对节点 first->next = third; second->next = tmp; if (tmp == NULL) break; first = second; // 前移到下一对节点 second = tmp; third = tmp->next; } return help->next; } };
[LeetCode]88. Swap Nodes in Pairs链表成对逆序
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原文地址：http://www.cnblogs.com/aprilcheny/p/4969766.html
