题面
题解
本题考查两路归并,我们可以用两个指针,新开一个虚拟节点,通过比较两个链表值的大小来移动指针,最后肯定有一个链表的指针到最后,然后将剩下一个链表的值复制即可
代码
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode *merge(ListNode *l1, ListNode *l2) {
ListNode *dummy = new ListNode(-1);
auto cur = dummy;
while (l1 && l2) {
if (l1->val < l2->val) {
cur->next = l1;
l1 = l1->next;
} else {
cur->next = l2;
l2 = l2->next;
}
cur = cur->next;
}
while (l1) cur->next = l1, l1 = l1->next, cur = cur->next;
while (l2) cur->next = l2, l2 = l2->next, cur = cur->next;
return dummy->next;
}
};