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。
链表1： 1->3->5->7
链表2： 2->4->6->8
结果： 1->2->3->4->5->6->7->8
思路
实现1-普通方法
/*
struct ListNode {
int val;
struct ListNode *next;
ListNode(int x) :
val(x), next(NULL) {
}
};*/
class Solution {
public:
ListNode* Merge(ListNode* pHead1, ListNode* pHead2)
{
if(!pHead1 || !pHead2)
return pHead1 ? pHead1 : pHead2;
ListNode* head = pHead1->val <= pHead2->val ? pHead1 : pHead2;
ListNode* p = head;
head == pHead1 ? pHead1 = pHead1->next : pHead2 = pHead2->next;
while(pHead1 && pHead2)
{
if(pHead1->val <= pHead2->val)
{
p->next = pHead1;
pHead1 = pHead1->next;
}
else
{
p->next = pHead2;
pHead2 = pHead2->next;
}
p = p->next;
}
p->next = pHead1 ? pHead1 : pHead2;
return head;
}
};
实现2-递归
/*
struct ListNode {
int val;
struct ListNode *next;
ListNode(int x) :
val(x), next(NULL) {
}
};*/
class Solution {
public:
ListNode* Merge(ListNode* pHead1, ListNode* pHead2)
{//对空指针的处理，也就是递归退出条件
if(!pHead1 || !pHead2)
return pHead1 ? pHead1 : pHead2;
//哪个小，就将其提取出来
if(pHead1->val < pHead2->val)
{
pHead1->next = Merge(pHead1->next, pHead2);
return pHead1;
}
else
{
pHead2->next = Merge(pHead1, pHead2->next);
return pHead2;
}
}
};
注意事项
- 当输入的是相等节点的链表
- 某一个链表为空，或者都为空链表
