题目:
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.
/**
* 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 && l2== NULL)
return NULL;
else if(l1 != NULL && l2 == NULL)
return l1;
else if(l1 == NULL && l2 != NULL)
return l2;
ListNode *head, *curr, *p1 = l1, *p2 = l2;
if(p1->val <= p2->val) {
head = p1;
p1 = p1->next;
}
else {
head = p2;
p2 = p2->next;
}
curr = head;
while(p1 != NULL && p2 != NULL) {
if(p1->val <= p2->val) {
curr->next = p1;
p1 = p1->next;
}
else {
curr->next = p2;
p2 = p2->next;
}
curr = curr->next;
}
while(p1 != NULL) {
curr->next = p1;
p1 = p1->next;
curr = curr->next;
}
while(p2 != NULL) {
curr->next = p2;
p2 = p2->next;
curr = curr->next;
}
return head;
}
};
本文介绍了一种合并两个已排序的单链表的方法,并通过拼接节点的方式返回一个新的排序链表。该算法首先检查输入链表是否为空,然后比较两链表的头节点值并进行连接,重复此过程直至遍历完所有节点。
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