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;
}
};