leetCode(10):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.

把小于x的链表单独取出组成新的链表，然后把两个链表相连。

/**
 * 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=head;
    	ListNode* preP=head;
    	ListNode* newHead=NULL;
    	ListNode* newP=newHead;
    	while(p)
    	{
    		if(p->val < x)
    		{
    			ListNode* tmp=p->next;
    			if(p==head)
    			{
    				head=tmp;
    				preP=tmp;
    			}
    			if(!newHead)
    			{
    				newHead=p;
    				newHead->next=NULL;
    				newP=newHead;
    			}
    			else
    			{
    				newP->next=p;
    				newP=newP->next;
    				newP->next=NULL;
    			}
    			if(preP!=tmp)
    				preP->next=tmp;
    			p=tmp;
    		}
    		else
    		{
    			if(preP!=p)
    				preP=preP->next;
    			p=p->next;
    		}
    	}
    	if(!newHead)
    		return head;
    	newP->next=head;
    	return newHead;
    }
};