今天看啥
热点:

LeeCode 连接两个有序单链表


 

题目:

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.

 

C代码:

 

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode* mergeTwoLists(struct ListNode* l1, struct ListNode* l2) {
    if(NULL == l1) return l2;
    if(NULL == l2) return l1;
    
    struct ListNode* head = NULL;
    
    if(l1->val < l2->val){
        head = l1;
        l1 = l1->next;
    }
    else{
        head = l2;
        l2 = l2->next;
    }
    
    struct ListNode* p = head;
    
    while(l1 && l2){
        if(l1->val < l2->val){
            p->next = l1;
            l1 = l1->next;
        }
        else{
            p->next = l2;
            l2 = l2->next;
        }
        p = p->next;
    }
    
    if(l1) p->next = l1;
    else p->next = l2;
    
    return head;
    
}


 

www.bkjia.comtruehttp://www.bkjia.com/cjjc/1003441.htmlTechArticleLeeCode 连接两个有序单链表 题目: 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...

相关文章

相关搜索: 两个

帮客评论

视觉看点