Code & Fun

第65天。

今天的题目是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.

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

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