LeetCode题解：Swap Nodes in Pairs

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.

思路：

一般来说是交换相邻的两个结点的内容就可以解决，无奈题目不允许这么做。只能交换指针了。要注意的是，指向每一对结点中前一个结点的指针也必须更改。

题解：

/** * 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)            return nullptr;            ListNode* nh = new ListNode(0);        nh->next = head;                ListNode* h0 = nh;        ListNode* h1 = head;        ListNode* h2 = head->next;                while(h2 != nullptr)        {            h1->next = h2->next;            h2->next = h1;            h0->next = h2;                        h0 = h1;            h1 = h1->next;            h2 = (h1 ? h1->next : nullptr);        }        h0 = nh->next;                delete nh;        return h0;    }};