【leetcode】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.

思路：

用两个指针，分别指向比x小的数，和其他的数，然后再整体地连起来，就OK了。注意最后得时候，要指向NULL.

/**
* Definition for singly-linked list.
* struct ListNode {
*     int val;
*     ListNode *next;
*     ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* partition(ListNode* head, int x) {
if(head==NULL || head->next==NULL) return head;

ListNode *p, *q,*temp=NULL,*record;
ListNode *newnode=(struct ListNode *)malloc(sizeof(struct ListNode));
ListNode *oldnode=(struct ListNode *)malloc(sizeof(struct ListNode));
record=oldnode;
q=newnode;
p=head;

while(p)
{
if(p->val >= x)
{
q->next=p;
q=q->next;
}
else
{
record->next=p;
record=record->next;
}
p=p->next;
}
q->next=NULL;
record->next=NULL;
record->next=newnode->next;
return oldnode->next;
}
};