Merge Two Sorted Lists
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.
合并两个有序链表
/**
* 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 *ans,*head;
if (l1==NULL && l2==NULL ) return NULL;
if (l1==NULL)
{
head=l2;
l2=l2->next;
}
else
if (l2==NULL)
{
head=l1;
l1=l1->next;
}
else
{
if (l1->val < l2->val)
{
head=l1;
l1=l1->next;
}
else
{
head=l2;
l2=l2->next;
}
}
ans=head;
while (l1!=NULL || l2!=NULL)
{
if (l1==NULL)
{
ans->next=l2;
ans=l2;
l2=l2->next;
}
else
if (l2==NULL)
{
ans->next=l1;
ans=l1;
l1=l1->next;
}
else
{
if (l1->val < l2->val)
{
ans->next=l1;
ans=l1;
l1=l1->next;
}
else
{
ans->next=l2;
ans=l2;
l2=l2->next;
}
}
}
return head;
}
};
Generate Parentheses
Generate Parentheses
Given n pairs of parentheses, write a function to generate all combinations of well-formed parentheses.
For example, given n = 3, a solution set is:
"((()))", "(()())", "(())()", "()(())", "()()()"
输出所有的括号匹配
class Solution
{
vector<string>ans;
public:
vector<string> generateParenthesis(int n)
{
string mark="";
dfs(mark,n,n);
return ans;
}
private:
void dfs(string mark,int x,int y)
{
if (x+y==0)
{
ans.push_back(mark);
return ;
}
if (x!=0)
dfs(mark+"(",x-1,y);
if (x<y)
dfs(mark+")",x,y-1);
}
};
Merge k Sorted Lists
Merge k sorted
linked lists and return it as one sorted list. Analyze and describe its complexity.
对K个有序链表排序
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class ListNodeCompare
{
public:
bool operator()(ListNode *l1,ListNode *l2)
{
return l1->val > l2->val;
}
};
class Solution
{
public:
ListNode* mergeKLists(vector<ListNode*>& lists)
{
ListNode *ans,*head,*p;
priority_queue<ListNode*,vector<ListNode*>,ListNodeCompare>q;
if (lists.size()==0) return NULL;
ans=(ListNode*)malloc(sizeof(ListNode));
ans->next=NULL;
head=ans;
for (int i=0;i<lists.size();i++)
if (lists[i]!=NULL)
q.push(lists[i]);
while (!q.empty())
{
p=q.top();
q.pop();
ans->next=p;
ans=ans->next;
if (p->next!=NULL)
q.push(p->next);
}
return head->next;
}
};
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.
对链表中每两个节点交换一下
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution
{
public:
ListNode* swapPairs(ListNode* head)
{
ListNode *p1,*pre,*p2;
if (head==NULL || head->next==NULL) return head;
p1=head;
pre=head;
while (1)
{
p2=p1->next;
p1->next=p2->next;
p2->next=p1;
if (pre==head)
head=p2;
else
pre->next=p2;
pre=p1;
p1=p1->next;
if (p1==NULL || p1->next==NULL) break;
}
return head;
}
};
Reverse Nodes in k-Group
Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.
If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is.
You may not alter the values in the nodes, only nodes itself may be changed.
Only constant memory is allowed.
For example,
Given this linked list: 1->2->3->4->5
For k = 2, you should return: 2->1->4->3->5
For k = 3, you should return: 3->2->1->4->5
对链表中每k个节点进行一次反转
class Solution
{
public:
ListNode* reverseKGroup(ListNode* head, int k)
{
int n=k;
int len=0;
ListNode *p=head;
if (head==NULL || head->next==NULL || k<=1) return head;
while (p)
{
len++;
p=p->next;
}
if (n>len) return head;
ListNode *q=head;
while (n--)
{
ListNode *ne=q->next;
q->next=p;
p=q;
q=ne;
}
if (len-k>=k)
head->next= reverseKGroup(q,k);
else
head->next=q;
return p;
}
};
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