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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.
其实关于链表的算法题思路都很easy,但是细节真的会崩溃,不管是反转链表,还是合并链表吗,还是删除指定节点,一定要小心临界条件的判断。就拿本题为例子吧,就是反转链表一个变形,它要求的是对相邻一对节点进行交换,交换的结束的条件怎么判断,(也就是怎么结束循环),如果交换的是头结点怎么处理,真心要小心小心再小心。上代码:
/** * Definition for singly-linked list. * struct ListNode { * int val; * struct ListNode *next; * }; */struct ListNode* swapPairs(struct ListNode* head) { struct ListNode* p = head; struct ListNode* pPre = NULL; struct ListNode* prePre = NULL; struct ListNode* pNext; while(p && p->next){ pPre = p; p=p->next; pNext = p->next; if(pPre == head) head = p; if(prePre) prePre->next = p; pPre->next = pNext; p->next = pPre; prePre = pPre; p = pNext; } return head;}
这里再说一个稍有难度的一个题,也是跟链表有关,不过这道题考察的重点是写一个堆排序的算法。
leetcode Merge k Sorted Lists
Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.别看题目就一句话描述,殊不知,越简短的描述越是个坑。
其实不想自己写堆排序直接用C++提供的make_heap(), push_heap(), pop_heap(), sort_heap()也能解决,但是C++提供的是大顶堆,而本题是要用到小顶堆。之前刷的时候对堆排序还不太了解,参考了: 这个博客讲的很详细。转载地址:http://fjtsi.baihongyu.com/