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。
/*
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==nullptr)
{
return pHead2;
}
if(pHead2==nullptr)
{
return pHead1;
}
ListNode* res=nullptr;
if(pHead1->val>=pHead2->val)
{
res=pHead2;
res->next=Merge(pHead1,pHead2->next);
}
else
{
res=pHead1;
res->next=Merge(pHead1->next,pHead2);
}
return res;
}
};
Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.
/**
* 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 dummy(-1);
ListNode *result = &dummy;
while(l1 && l2)
{
if(l1->val <= l2->val)
{
result->next = l1;
result = result->next;
l1 = l1->next;
}
else
{
result->next = l2;
result = result->next;
l2 = l2->next;
}
}
if(l1)
result->next = l1;
else if(l2)
result->next = l2;
return dummy.next;
}
ListNode *mergeKLists(vector<ListNode *> &lists)
{
if (lists.empty()) return NULL;
int len = lists.size();
while (len > 1)
{
for (int i = 0; i < len / 2; ++i)
{
lists[i] = mergeTwoLists(lists[i], lists[len - 1 - i]);
}
len = (len + 1) / 2;
}
return lists.front();
}
};