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.

用了哨兵之后，能够回避很多问题。在此需要注意的问题是，对于链表最后一个元素，应该考虑他是否大于等于x。之前偷懒没有处理这个元素，但是题意要求保持原顺序。

/**
 * 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) {
        ListNode *gard = new ListNode(0);
        ListNode *par = new ListNode(x);
        ListNode *pp = par;
        gard->next = head;
        ListNode *pre = gard;
        
        ListNode *temp = head;
        if(head == NULL) return head;
        
        while(temp -> next != NULL)
        {
            if(temp -> val >=x)
            {
                pre -> next = temp ->next;
                temp ->next = NULL;
                pp->next = temp;
                pp = pp->next;
                temp =pre->next;
            }
            else
            {
                pre =pre->next;
                temp = temp->next;
            }
        }
        if(temp -> val >=x)
        {
            pp->next = temp;
            pre->next = NULL;
            temp =pre;
        }
        temp ->next = par ->next;
        
        return gard->next;
    }
};