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.

分成两段，按照x，最后连接起来。 看了别人的答案，突然发现safeguard写成object instance更方便一些。

struct ListNode {
	int val;
	ListNode *next;
	ListNode(int x) :
			val(x), next(NULL) {
	}
};

class Solution {
public:
    ListNode *partition(ListNode *head, int x) {
        ListNode head1(0),head2(0);
        ListNode *node1=&head1, *node2=&head2;
        while(head){
        	if (head->val<x)
        		node1=node1->next=head;
        	else
        		node2=node2->next=head;
        	head=head->next;
        }
        node2->next=NULL;
        node1->next=head2.next;
        return head1.next;
    }
};