leetcode：Partition List

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：

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/**  * 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;     } };