[LeetCode]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.

这道题要求把链表中小于x的结点顺序放到链表的前面，同时不改变链表中结点原来的相对位置。

仍然使用链表中最常用的双指针法。为了方便操作，增设一个头结点。
1. 先定位到第一个不小于x的结点，pre指向该节点的前驱。
2. 然后从该节点的下一个结点开始，若当前结点的值小于x，则将该结点移动到pre后面，否则继续向后遍历。
3. 将边界值的情况考虑清楚即可。

/**
 * 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){
            ListNode* pre = new ListNode(0);
            pre->next = head;
            head = pre;
            while (pre->next&&pre->next->val < x){
                pre = pre->next;
            }
            ListNode* p = pre->next;
            if (p){
                while (p->next){
                    ListNode* r = p->next;
                    if (p->next->val < x){
                        p->next = r->next;
                        r->next = pre->next;
                        pre->next = r;
                        pre = r;
                    }
                    else{
                        p = r;
                    }

                }
            }
            return head->next;
        }
        return head;
    }
};