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.
题目意思为合并两个链表,不能用额外的空间(新链由两个链拼接而成)
思路:
比较简单,直接定义两个指针,分别指向两条链,比较所指向节点的大小 如果指针指向的值为小的,那么就让新链指向小的,同时该指针往后移动
,新链的当前指针也往后移动。直到两条链中有一条遍历到了结尾,然后将剩余的链拼接到新链
代码如下:
<span style="font-size:18px;">/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) { if(l1==NULL) return l2; if(l2==NULL) return l1; ListNode *p,*q; p=l1; q=l2; ListNode *result=new ListNode(0); ListNode *temp =result; while(p&&q) { if(p->val<q->val) { temp->next=p; temp=p; p=p->next; } else { temp->next=q; temp=q; q=q->next; } } if(p) { temp->next=p; } if(q) { temp->next=q; } return result->next; } };</span>
时间: 2025-01-04 16:40:25