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.
把链表分段,小于x的放在前面,其他的放后面。
class Solution {
public:
ListNode* partition(ListNode* head, int x) {
if(head==NULL)return NULL;
ListNode* p=new ListNode(0);
ListNode* phead=p;
ListNode* q=new ListNode(0);
ListNode* qhead=q;
while(head){
if(head->val<x){
p->next=head;
p=p->next;
}else{
q->next=head;
q=q->next;
}
head=head->next;
}
p->next=qhead->next;
q->next=NULL;
return phead->next;
}
};