Swap Nodes in Pairs leetcode

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.

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未优化的代码，用3个临时指针

ListNode* swapPairs(ListNode* head) {
if (head == nullptr || head->next == nullptr)
return head;
ListNode *pre = head;
ListNode *cur = pre->next;
ListNode *net = cur->next;
head = cur;
while (true)
{
cur->next = pre;
if (net == nullptr || net->next == nullptr) {
pre->next = net;
break;
}
pre->next = net->next;
pre = net;
cur = pre->next;
net = cur->next;
}
return head;
}

惊人的解法，用双重指针，开阔思路，本以为自己可以随心所欲使用双指针，看来还是差得远

ListNode *swapPairs(ListNode *head) {
ListNode **p = &head;

while (*p && (*p)->next) {
ListNode *t = (*p)->next;

(*p)->next = t->next;
t->next = *p;
*p = t;

p = &(*p)->next->next;
}

return head;
}