acttell的博客

Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity. 思路：建一个大小为K的小顶堆（优先队列默认大顶堆，需要自定义降序，变成小顶堆)，先把K个链表的头结点扔进堆，取了堆顶元素后，接着把堆顶元素所在链表的非空结点扔进堆。 #include <iost...

http://www.cnblogs.com/grandyang/p/4355505.html

https://www.cnblogs.com/dancingrain/p/3405197.html

https://blog.youkuaiyun.com/lks1139230294/article/details/52775517

https://www.cnblogs.com/DarrenChan/p/8764608.html

http://www.cnblogs.com/grandyang/p/4249905.html

You are given two linked lists representing two non-negative numbers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a l...

思路： 如果两个单向链表l1 l2有公共结点， 那么l1 l2从某一结点开始，它们next指向公共结点。 从公共结点开始后面所有结点均重合。 分别遍历l1 l2得到它们的长度n1 n2，并求出|n1-n2|。 在长链表上先遍历|n1-n2|，再同步遍历l1 l2，直到找到相同的结点。 时间复杂度为O(n1+n2)。...

思路：3个辅助指针,不断移动指针位置 #include <iostream> using namespace std; typedef struct Node { int val; struct Node* next; }node; node* reverse(node* head) { node* p=NULL; node* q=NULL; ...

题目：链表A和B都升序，合并为一个升序的链表。 思路：递归 #include <iostream> using namespace std; typedef struct Node { int val; struct Node* next; }node; node* merge(node* a,node* b) { if(a==NULL) r...

https://blog.youkuaiyun.com/HelloZEX/article/details/81126071?utm_source=blogxgwz6