剑指offer：合并两个排序的链表

我是真的傻逼。
https://www.acwing.com/solution/content/744/

/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */
class Solution {
    
public:
    ListNode* merge(ListNode* l1, ListNode* l2) {
    
        // 我在刚开始也想到弄一个虚的头节点来着！
        ListNode* dummy = new ListNode(0);
        ListNode* p = dummy;
        
        ListNode* i = l1;
        ListNode* j = l2;

        while(i!=NULL && j!=NULL)
        {
    
            if(i->val <= j->val)
            {
    
                p->next = i;
                i = i->next;
                p = p->next;
            }
            else
            {
    
                p->next = j;
                j = j->next;
                p = p->next;
            }
        }
        
        p -> next = (i != NULL ? i : j);
        return dummy -> next;
    }
};

看看y总的多简洁：

/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */
class Solution {
    
public:
    ListNode* merge(ListNode* l1, ListNode* l2) {
    
        ListNode *dummy = new ListNode(0);
        ListNode *cur = dummy;
        while (l1 != NULL && l2 != NULL) {
    
            if (l1 -> val < l2 -> val) {
    
                cur -> next = l1;
                l1 = l1 -> next;
            }
            else {
    
                cur -> next = l2;
                l2 = l2 -> next;
            }
            cur = cur -> next;
        }
        cur -> next = (l1 != NULL ? l1 : l2);
        return dummy -> next;
    }
};

作者：yxc
链接：https://www.acwing.com/solution/content/744/
来源：AcWing
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