leetcode

题目：Merge Two Sorted Lists

Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.

个人思路：

1、合并两个有序链表，一般就是采用归并的方法，分别设置一个指针用于遍历链表，然后挨个比较节点大小，将小的那个节点作为新链表的下一个节点，然后让小节点的指针向后走一步，直到某个链表走完

2、当某个链表走完，则将另外一个链表的剩余部分接到新链表的后面就行了

代码：

#include <stddef.h>

struct ListNode
{
    int val;
    ListNode *next;
    ListNode(int x) : val(x), next(NULL) {};
};

class Solution
{
public:
    ListNode* mergeTwoLists(ListNode *l1, ListNode *l2)
    {
        //若l1或者l2为空
        if (!l1)
        {
            return l2;
        }
        if (!l2)
        {
            return l1;
        }

        ListNode *cur_l1 = l1;
        ListNode *cur_l2 = l2;
        ListNode *head_new = NULL;
        ListNode *cur_new = NULL;

        while (cur_l1 && cur_l2)
        {
            if (cur_l1->val <= cur_l2->val)
            {
                if (!cur_new)
                {
                    head_new = cur_new = cur_l1;
                }
                else
                {
                    cur_new->next = cur_l1;
                    cur_new = cur_new->next;
                }
                cur_l1 = cur_l1->next;
            }
            else
            {
                if (!cur_new)
                {
                    head_new = cur_new = cur_l2;
                }
                else
                {
                    cur_new->next = cur_l2;
                    cur_new = cur_new->next;
                }
                cur_l2 = cur_l2->next;
            }
        }

        if (!cur_l1)
        {
            while (cur_l2)
            {
                cur_new->next = cur_l2;
                cur_new = cur_new->next;
                cur_l2 = cur_l2->next;
            }
        }

        if (!cur_l2)
        {
            while (cur_l1)
            {
                cur_new->next = cur_l1;
                cur_new = cur_new->next;
                cur_l1 = cur_l1->next;
            }
        }

        return head_new;
    }
};

int main()
{
    return 0;
}

网上基本都是这个归并的思路，属于比较基础的题目。