Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.
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class Solution {
public:
ListNode* (ListNode* l1, ListNode* l2) {
if(l1==NULL)
{
return l2;
}
if(l2==NULL)
{
return l1;
}
ListNode* head=(ListNode*)malloc(sizeof(ListNode));
ListNode* p=head;
ListNode* p1=l1;
ListNode* p2=l2;
while(p1!=NULL&&p2!=NULL)
{
if(p1->val<p2->val)
{
p->next=p1;
p=p->next;
p1=p1->next;
}
else
{
p->next=p2;
p=p->next;
p2=p2->next;
}
}
if(p1==NULL)
{
p->next=p2;
}
else
{
p->next=p1;
}
return head->next;
}
};
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Swap Nodes in Pairs
Given a linked list, swap every two adjacent nodes and return its head.
For example,
Given 1->2->3->4, you should return the list as 2->1->4->3.
Your algorithm should use only constant space. You may not modify the values in the list, only nodes itself can be changed.
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class Solution {
public:
ListNode* swapPairs(ListNode* head) {
if (head == NULL || head->next == NULL)
{
return head;
}
ListNode* p1 = head;
ListNode* p2 = head->next;
ListNode* pre = (ListNode*)malloc(sizeof(ListNode));
pre->next = p1;
ListNode* newHead = head->next;
while (p1 != NULL&&p2 != NULL)
{
pre->next = p2;
p1->next = p2->next;
p2->next = p1;
pre = p1;
p1 = p1->next;
if (pre->next != NULL&&p1->next!=NULL)
{
p2 = p1->next;
}
else
{
break;
}
}
return newHead;
}
};
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