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题目描述：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.
分析：
题目中说不能直接改变值，所以只能改变结点。
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 == nullptr || head->next == nullptr) return head; ListNode dummy(-1); dummy.next = head; for(ListNode *prev = &dummy, *cur = prev->next, *next = cur->next; next; prev = cur, cur = cur->next, next = cur ? cur->next: nullptr) { prev->next = next; cur->next = next->next; next->next = cur; } return dummy.next; } };
LeetCode 024 Swap Nodes in Pairs
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原文地址：http://www.cnblogs.com/510602159-Yano/p/4278962.html
