问题描述：
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.
问题分析：
该题主要是将两个有序的链表合并为一个有序的列表，方法很简单，直接看代码即可
/**
* 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) {
ListNode dummy(INT_MIN); //先创建一个小的头结点
ListNode *tail = &dummy;
while(l1 && l2)
{
if(l1->val < l2->val)
{
tail->next = l1;
l1 = l1->next;
}
else
{
tail->next = l2;
l2 = l2->next;
}
tail = tail->next;
}
tail->next = l1 ? l1 : l2;
return dummy.next;
}
};
转载请注明原文地址: https://ju.6miu.com/read-10736.html
