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一天一道LeetCode系列 ##(一)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. ##(二)解题 这题是剑指offer上的老题了,剑指上面用的是递归,我写了个非递归的版本。
/**
* 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* head = NULL;
if(l1==NULL && l2==NULL) return head;
ListNode* p = NULL;
ListNode* p1 = l1;
ListNode* p2 = l2;
while(p1 != NULL || p2!=NULL)
{
if(p1 != NULL && p2 != NULL)
{
if(p1->val < p2->val)
{
if(head == NULL) //记录头节点head
{
head = p1;
p = head;
}
else {p->next = p1;p=p->next;}
p1=p1->next;
}
else
{
if(head == NULL)
{
head = p2;
p = head;
}
else {p->next = p2;p=p->next;}
p2=p2->next;
}
}
if(p1 == NULL && p2 != NULL)
{
if(head == NULL)
{
head = p2;
p = head;
}
else {p->next = p2;p=p->next;}
p2=p2->next;
}
if(p1 != NULL && p2 == NULL)
{
if(head == NULL)
{
head = p1;
p = head;
}
else {p->next = p1;p=p->next;}
p1=p1->next;
}
}
return head;
}
};
递归版本:
class Solution {
public:
ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
if(l1 == NULL) return l2;
if(l2 == NULL) return l1;
ListNode* head;
if(l1->val < l2->val)
{
head = l1;
head->next = mergeTwoLists(l1->next , l2);
}
else
{
head = l2;
head->next = mergeTwoLists(l1 , l2->next);
}
return head;
}
};