Code & Func

第65天。

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.

和之前做的链表翻转有点像，可以用指针的指针来做：

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bool swap(ListNode **head) {
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    if (*head == nullptr || ( *head)->next == nullptr) return false;
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    ListNode *p = *head;
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    ListNode *next = p->next;
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    p->next = next->next;
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    next->next = p;
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    *head = next;
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    return true;
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}
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ListNode* swapPairs(ListNode* head) {
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    ListNode **p = &head;
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    while(swap(p))
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        p=&((*p)->next->next);
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    return head;
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}