将两个有序链表合并为一个新的有序链表并返回。新链表是通过拼接给定的两个链表的所有节点组成的。
示例:
输入:1->2->4, 1->3->4
输出:1->1->2->3->4->4
C_非递归合并
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* struct ListNode *next;
* };
*/
struct ListNode* mergeTwoLists(struct ListNode* l1, struct ListNode* l2){
if(l1==NULL)
return l2;
if(l2==NULL)
return l1;
struct ListNode* newNode;
struct ListNode* tail;
if(l1->val<l2->val){
newNode=l1;
l1=l1->next;
}
else{
newNode=l2;
l2=l2->next;
}
tail=newNode;
while(l1&&l2){
if(l1->val<l2->val){
tail->next=l1;
l1=l1->next;
}
else{
tail->next=l2;
l2=l2->next;
}
tail=tail->next;
}
if(l1)
tail->next=l1;
else if(l2)
tail->next=l2;
return newNode;
}
C_递归合并
struct ListNode* mergeTwoLists(struct ListNode* l1, struct ListNode* l2) {
struct ListNode* newNode;
if(!l1)
return l2;
if(!l2)
return l1;
if(l1->val<l2->val)
{
newNode=l1;
newNode->next=mergeTwoLists(l1->next,l2);
}
else
{
newNode=l2;
newNode->next=mergeTwoLists(l1,l2->next);
}
return newNode;
}
C++
class Solution {
public:
ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
if(l1 != NULL && l2 != NULL) {
ListNode *p = l1;
ListNode *q = l2;
ListNode *t, *h;//t为新链表的连接指针,h为新链表的头指针
if(p->val > q->val) {
t = q;
h = q;
q = q->next;
} else {
t = p;
h = p;
p = p->next;
}
while(p != NULL && q != NULL) {
if(p->val > q->val) {
t->next = q;
t = t->next;
q = q->next;
} else {
t->next = p;
t = t->next;
p = p->next;
}
}
while(p != NULL && q == NULL) {
t->next = p;
p = p->next;
t = t->next;
}
while(p == NULL && q != NULL) {
t->next = q;
q = q->next;
t = t->next;
}
while(p == NULL && q == NULL) {
return h;
}
}
if(l1 == NULL && l2 != NULL) {
return l2;
}
if(l1 != NULL && l2 == NULL) {
return l1;
}
if(l1 == NULL && l2 == NULL) {
return NULL;
}
}
};