题目：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.

分析：

--->题意：给定一个单链表和一个x，把链表中小于x的放到前面，大于等于x的放到后面，每部分元素的原始相对位置不变。

--->思路：遍历一遍链表，把小于x的都挂到part1后，把大于等于x的都放到part2后，最后再把大于等于的链表挂到小于链表的后面就可以了。

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) {        ListNode *part1 = new ListNode(0), *part2 = new ListNode(0), *node1, *node2;        node1 = part1;        node2 = part2;        while (head) {            if (head->val < x) {                node1->next = head;                node1 = node1->next;            } else {                node2->next = head;                node2 = node2->next;            }            head = head->next;        }        node2->next = NULL;        node1->next = part2->next;        return part1->next;    }};