LKJZ25-合并两个排序的链表

https://leetcode-cn.com/problems/he-bing-liang-ge-pai-xu-de-lian-biao-lcof/

 

取小的尾插

1.哨兵位的头节点

class Solution {
public:
    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
        if(l1==nullptr){
            return l2;
        }
        if(l2==nullptr){
            return l1;
        }
        ListNode*head,*tail;
        head=tail=new ListNode;
        while(l1!=nullptr&&l2!=nullptr){
            if(l1->val<l2->val){
                tail->next=l1;
                l1=l1->next;
                tail=tail->next;
            }
            else{
                tail->next=l2;
                l2=l2->next;
                tail=tail->next;
            }
        }
        while(l1!=nullptr){
            tail->next=l1;
            l1=l1->next;
            tail=tail->next;
        }
        while(l2!=nullptr){
            tail->next=l2;
            l2=l2->next;
            tail=tail->next;
        }
        ListNode*ret=head->next;
        delete head;
        return ret;
    }
};

 

2.先拿一个小的节点下来赋给head和tail

class Solution {
public:
    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
        if(l1==nullptr){
            return l2;
        }
        if(l2==nullptr){
            return l1;
        }
        ListNode*head=nullptr,*tail=nullptr;
        if(l1->val<l2->val){
            head=tail=l1;
            l1=l1->next;
        }
        else{
            head=tail=l2;
            l2=l2->next;
        }
        while(l1!=nullptr&&l2!=nullptr){
            if(l1->val<l2->val){
                tail->next=l1;
                l1=l1->next;
                tail=tail->next;
            }
            else{
                tail->next=l2;
                l2=l2->next;
                tail=tail->next;
            }
        }
        while(l1!=nullptr){
            tail->next=l1;
            l1=l1->next;
            tail=tail->next;
        }
        while(l2!=nullptr){
            tail->next=l2;
            l2=l2->next;
            tail=tail->next;
        }
        return head;
    }
};