86. 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.

题目含义：所有小于X的元素都排在大于等于X的前面。

思路：创建两个链表，分别连接比x大的和小的数，最后构成一个整体

    public ListNode partition(ListNode head, int x) {
        ListNode small = new ListNode(0);
        ListNode big = new ListNode(0);
        ListNode smallCur = small;
        ListNode bigCur = big;
        while (head != null)
        {
            if (head.val<x)
            {
                smallCur.next = head;
                smallCur = head;
            }else
            {
                bigCur.next =head;
                bigCur = head;
            }
            head=head.next;
        }
        bigCur.next = null;
        smallCur.next = big.next;
        return small.next;        
    }