题目描述
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.
思路
要把两个已经排序的list(L1, L2)合并并且排好序。那么首先比较的就是两个队列的head,然后将value较小的head作为新list的head。
这里我们假设是L1->head成为了newlist->head,所以newlist->val=L1->value,然后L1=L1->next,但是L2不变。继续比较L1和L2的value值大小。
这里假设L2->val< L1->val,那么newlist->val=L2->val,然后L2=L2->next,L1不变。
重复操作知道L1,L2遍历完成。
class Solution {
public:
ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
if (l1 == NULL) return l2;
if (l2 == NULL) return l1;
ListNode* head = NULL;
if (l1->val < l2->val) {
head = l1;
l1 = l1->next;
} else {
head = l2;
l2 = l2->next;
}
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;
}
};