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.

思路：分别用两个链表，一个保存比指定数小的元素，一个保存大于等于指定数的元素，最后把第二个链表接在第一个链表之后即可。

实现代码：

/**
 * 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 node1(0), node2(0);
        ListNode *p1 = &node1, *p2 = &node2;
        while (head) {
            if (head->val < x)
                p1 = p1->next = head;
            else
                p2 = p2->next = head;
            head = head->next;
    }
        p2->next = NULL;
        p1->next = node2.next;
        return node1.next;
    }
};