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Merge Two Sorted Lists
/**
* 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;
}
};
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:
"((()))", "(()())", "(())()", "()(())", "()()()"
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);
}
};/**
* 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;
}
};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.
/**
* 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;
}
};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
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;
}
};版权声明：本文为博主原创文章，未经博主允许不得转载。
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原文地址：http://blog.csdn.net/u011932355/article/details/47658597
