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.
比起第一次写简洁多了。
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 if (head == NULL) return head; 13 ListNode h(0), *first = head, *second, *prev = &h, *nextFirst; 14 h.next = head; 15 16 while (first) { 17 second = first->next; 18 if (!second) break; 19 prev->next = second; 20 nextFirst = second->next; 21 second->next = first; 22 first->next = nextFirst; 23 prev = first; 24 first = nextFirst; 25 } 26 27 return h.next; 28 } 29 };
Leetcode | Swap Nodes in Pairs
原文：http://www.cnblogs.com/linyx/p/4036498.html