【数据结构_链表_最小堆】 链表找环，链表找交点，合并有序链表

Linked List Cycle

Given a linked list, determine if it has a cycle in it.

Follow up:
Can you solve it without using extra space?

对链表反序： while(p) { q=p->next; p->next=f; f=p;p=q;}

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public:    bool hasCycle(ListNode *head) {        ListNode *p=head,*q,*f=NULL;        while(p){//对链表反序，如果p回到起点，则有环            q=p->next;            p->next=f;            f=p;p=q;            if(p==head) return true;//有环        }        return false;//沒有环，p到了NULL    }};

合并有序链表

用优先队列（小顶堆）

class Solution {public:    struct cmp{        bool operator () ( const pair<int, int> &a, const pair<int, int > &b){            return a.first >= b.first;        }    }cc;    ListNode *mergeKLists(vector<ListNode *> &lists) {        priority_queue<pair<int, int>, vector<pair<int, int> >, cmp> que;        for(int i=0; i<lists.size(); i++) if(lists[i])             que.push(pair<int, int>(lists[i]->val, i));                ListNode *head =NULL, *tail =NULL;        while(!que.empty()){            pair<int, int> t = que.top(); que.pop();            int i = t.second;            if(!head) head = lists[i];            else tail -> next = lists[i];            tail = lists[i];            lists[i] = lists[i]->next;            if(lists[i])                 que.push(pair<int, int>(lists[i]->val, i));        }                if(tail) tail->next = NULL;                return head;    }};