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：很容易想到的方式是递归，时间复杂度递推公式为T(n) = T(n-2) + 2，可知时间复杂度为O(n)。代码：

class Solution {
public:
ListNode *swapPairs(ListNode *head) {
if(head == NULL || head->next == NULL) return head;

ListNode * tail = swapPairs(head->next->next);
ListNode * tmp = head->next;
head->next = tail;
tmp->next = head;

return tmp;
}
};

迭代版：

class Solution {
public:
ListNode *swapPairs(ListNode *head) {
if(head == NULL || head->next == NULL) return head;

ListNode * dummy = new ListNode(-1);
dummy->next = head;

for(ListNode * pre_p = dummy, *p = head; p && p->next;){
ListNode * tmp = p->next;
p->next = p->next->next;
tmp->next = p;
pre_p->next = tmp;

pre_p = p;
p = p->next;
}

return dummy->next;
}
};