问题描述：
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.
示例： Given 1->4->3->2->5->2 and x = 3, return 1->2->2->4->3->5.
问题分析：
该类问题是节点删除和添加的综合应用问题，解题时注意增删节点的细节即可。
过程详见代码：
/** * 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) return head; ListNode* first = NULL, *last; ListNode res(0); res.next = head; last = &res; while (head) { if (head->val >= x) { if (first == NULL) { first = last; } } else { if (first) { last->next = head->next; head->next = first->next; first->next = head; first = first->next; head = last; } } last = head; head = head->next; } return res.next; } };