题目描述
输入两个单调递增的链表,输出两个链表合成后的链表,当然我们需要合成后的链表满足单调不减规则。
(1)递归版本
/* struct ListNode { int val; struct ListNode *next; ListNode(int x) : val(x), next(NULL) { } };*/
class Solution
{
public:
ListNode* Merge(ListNode* pHead1, ListNode* pHead2)
{
if(nullptr == pHead1)
return pHead2;
if(nullptr == pHead2)
return pHead1;
if(pHead1->val > pHead2->val)
{
pHead2->next = Merge(pHead1,pHead2->next);
return pHead2;
}
else
{
pHead1->next = Merge(pHead1->next,pHead2);
return pHead1;
}
}
};
(2)非递归版本
/* struct ListNode { int val; struct ListNode *next; ListNode(int x) : val(x), next(NULL) { } };*/
class Solution
{
public:
ListNode* Merge(ListNode* pHead1, ListNode* pHead2)
{
//非递归
if(nullptr == pHead1 && nullptr == pHead2)
return nullptr;
ListNode * pHead = new ListNode(0);
ListNode * cur = pHead ;
while(true)
{
if(nullptr == pHead1)
{
cur->next =pHead2;
break;
}
else if(nullptr == pHead2)
{
cur->next = pHead1;
break;
}
if(pHead1->val > pHead2->val)
swap(pHead1, pHead2);
cur->next = pHead1;
cur = cur->next;
pHead1 = pHead1->next;
}
return pHead->next;
}
};
<stron>:合并两个排序的链表</stron>