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;}