每日刷题札记–合并两个有序链表
方法一:建立新链表，依次按升序链接两个链表的元素
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode() : val(0), next(nullptr) {}
* ListNode(int x) : val(x), next(nullptr) {}
* ListNode(int x, ListNode *next) : val(x), next(next) {}
* };
*/
class Solution {
public:
ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
struct ListNode dummy;
ListNode* pre = &dummy;
while(l1 && l2)
{
if(l1->val < l2->val)
{
pre->next = l1;
l1 = l1->next;
}
else
{
pre->next = l2;
l2 = l2->next;
}
pre = pre->next;
}
pre->next = l1?l1:l2;
return dummy.next;
}
};
方法二:递归
图片来源于Leetcode官方题解
class Solution {
public:
ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
if (l1 == nullptr) {
return l2;
} else if (l2 == nullptr) {
return l1;
} else if (l1->val < l2->val) {
l1->next = mergeTwoLists(l1->next, l2);
return l1;
} else {
l2->next = mergeTwoLists(l1, l2->next);
return l2;
}
}
};