题单：LeetCode链表

2. 两数相加

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        auto dummy = new ListNode(-1), cur = dummy;
        int t = 0;
        while(l1 || l2 || t)
        {
            if(l1) t += l1->val, l1 = l1->next;
            if(l2) t += l2->val, l2 = l2->next;
            cur = cur->next = new ListNode(t % 10);
            t /= 10;
        }
        return dummy->next;
    }
};

作者：NFYD
链接：https://www.acwing.com/activity/content/code/content/1437840/
来源：AcWing
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。

19. 删除链表的倒数第 N 个结点

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int k) {
        //建立一个虚拟头结点，因为头结点可能会被删掉，这样便于操作
        auto dummy = new ListNode(-1);
        dummy->next = head;

        int len = 0; //计算链表长度
        for(auto p = dummy; p; p = p->next) len ++ ;

        auto p = dummy;
        //跳到要删除节点的前一个点，然后p->next = p->next->next;
        for(int i = 0; i < len - k - 1; i ++ ) p = p->next;
        p->next = p->next->next;

        return dummy->next;
    }
};

作者：NFYD
链接：https://www.acwing.com/activity/content/code/content/3860044/
来源：AcWing
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。

21. 合并两个有序链表

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
        auto dummy = new ListNode(-1), tail = dummy;
        while(l1 && l2)
        {
            if(l1->val < l2->val)
            {
                tail = tail->next = l1;
                l1 = l1->next;
            }
            else
            {
                tail = tail->next = l2;
                l2 = l2->next;
            }
        }
        if(l1) tail->next = l1;
        if(l2) tail->next = l2;
        return dummy->next;
    }
};

23. 合并K个升序链表

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    struct Cmp
    {
        bool operator() (ListNode* a, ListNode* b)
        {
            return a->val > b->val;
        }
    };
    ListNode* mergeKLists(vector<ListNode*>& lists) {
        priority_queue<ListNode*, vector<ListNode*>, Cmp> heap;
        auto dummy = new ListNode(-1);
        auto tail = dummy;
        for(auto l : lists) if(l) heap.push(l);
        while(heap.size())
        {
            auto t = heap.top();
            heap.pop();

            tail = tail->next = t;
            if(t->next) heap.push(t->next);
        }
        return dummy->next;
    }
};