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：

/**
 * 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;
    }
};