leetcode[86]:Partition List

Partition List

Given a linked list and a value x, partition it such that all nodes less than x come before nodes greater than or equal to x.

You should preserve the original relative order of the nodes in each of the two partitions.

For example,

Given 1->4->3->2->5->2 and x = 3,

return 1->2->2->4->3->5.

/**
* Definition for singly-linked list.
* struct ListNode {
*     int val;
*     struct ListNode *next;
* };
*/
struct ListNode* partition(struct ListNode* head, int x) {
int flag=0;
struct ListNode* tail,*tmp,*head1,*tmp2,*tail1;
if(!head) return NULL;
if(head->val<x)
{
head1=head;
tmp=head->next;
head1->next=NULL;
flag=10;
}
else
{
tail=head;
tail1=head;
flag=1;
tmp=head->next;
tail1->next=NULL;
}

while(tmp)
{
if(tmp->val<x)
{
if(flag==10)
{
head1->next=tmp;
head1=head1->next;
tmp=tmp->next;

}
else if(flag==1)
{
head1=tmp;
tmp2=tmp->next;
head1->next=tail;
tmp=tmp2;
head=head1;
flag=11;

}
else
{
tmp2=tmp->next;
head1->next=tmp;
head1=head1->next;
head1->next=tail;
tmp=tmp2;

}
}
else
{
if(flag==10)
{
tmp2=tmp->next;
tmp->next=NULL;
tail=tmp;
tail1=tail;
head1->next=tail1;
tmp=tmp2;
flag=11;

}
else if(flag==01)
{
tmp2=tmp->next;
tmp->next=NULL;
tail1->next=tmp;
tail1=tail1->next;
tmp=tmp2;
}
else
{
tmp2=tmp->next;
tail1->next=tmp;
tail1=tail1->next;
tail1->next=NULL;
tmp=tmp2;
}
}

}
return head;
}

head：小数头指针

tail：大数头指针

head1：小数头移指针

tail1：大数头移动指针

flag：01小无头大有头（head未赋值，tail已赋值）

10小有头大无头（head已赋值，tail未赋值）

11小有头大有头（head已赋值，tail已赋值）

做的很复杂，运行时间也很长12ms，需要改进算法。

改进后：

/**
* Definition for singly-linked list.
* struct ListNode {
*     int val;
*     struct ListNode *next;
* };
*/
struct ListNode* partition(struct ListNode* head, int x) {
struct ListNode *head0,*head1,*tmp,*tmp1,*tmp2,*tmp0,*tape=0;
if(!head) return NULL;
if(!head->next) return head;
head0=(struct ListNode *)malloc(sizeof(struct ListNode *));
head0->val=0;
head0->next=NULL;
head1=(struct ListNode *)malloc(sizeof(struct ListNode *));
head1->val=0;
head1->next=NULL;
tmp1=head0;
tmp2=head1;
while(head)
{
if(head->val<x)
{
tmp0=head->next;
head->next=NULL;
tmp1->next=head;
tmp1=tmp1->next;
head=tmp0;
}
else
{
tmp0=head->next;
head->next=NULL;
tmp2->next=head;
tmp2=tmp2->next;
head=tmp0;
}
}
tmp1->next=head1->next;
return head0->next;

}

两个虚头链表，一个放小数，一个放大数。去头合并即可。