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牛客网链接：合并两个排序的链表
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。
递归代码：
/*
struct ListNode {
int val;
struct ListNode *next;
ListNode(int x) :
val(x), next(NULL) {
}
};*/
class Solution {
public:
ListNode* Merge(ListNode* head1, ListNode* head2)
{
if (head1 == nullptr)
return head2;
else if (head2 == nullptr)
return head1;
ListNode* pMergeHead = nullptr;
if(head1->val > head2->val)
{
pMergeHead = head2;
pMergeHead->next = Merge(head1,head2->next);
}
else
{
pMergeHead = head1;
pMergeHead->next = Merge(head1->next,head2);
}
return pMergeHead;
}
};
非递归代码
/*
struct ListNode {
int val;
struct ListNode *next;
ListNode(int x) :
val(x), next(NULL) {
}
};
*/
class Solution {
public:
ListNode* Merge(ListNode* list1, ListNode* list2)
{
if(list1 == nullptr)
return list2;
if(list2 == nullptr)
return list1;
ListNode *pMergeHead = nullptr;
ListNode *current = nullptr;
while(list1!=nullptr && list2 != nullptr)
{
if(list1->val <= list2->val)
{
if(pMergeHead == nullptr)
{
current = pMergeHead = list1;
}
else
{
current->next = list1;
current = current->next;
}
list1 = list1->next;
}
else
{
if(pMergeHead == nullptr)
{
current = pMergeHead = list2;
}
else
{
current->next = list2;
current = current->next;
}
list2 = list2->next;
}
}
if(list1 == nullptr)
{
current->next = list2;
}
if(list2 == nullptr)
{
current->next = list1;
}
return pMergeHead;
}
};