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.

##Approach 1 解题思路为one pass，采用two pointers，一个连接小的部分，一个连接大的部分，再将两部分连接。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode partition(ListNode head, int x) {
        ListNode smallerHead = new ListNode(0);
        ListNode biggerHead = new ListNode(0);
        ListNode smaller = smallerHead, bigger = biggerHead;
        while(head != null) {
            if(head.val < x) {
                smaller.next = head;
                smaller = smaller.next;
            } else {
                bigger.next = head;
                bigger = bigger.next;
            }
            head = head.next;
        }
        smaller.next = biggerHead.next;
        bigger.next = null;
        return smallerHead.next;
    }
}

##Approach 2 记住小的那部分的尾节点，one pass链表，凡遇到小的节点，就把它从当前位置删除，插入到小的部分的尾节点之后。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode partition(ListNode head, int x) {
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode small = dummy;
        ListNode pre = dummy;
        ListNode cur = head;
        while(cur != null) {
            if(cur.val < x) {
                if(pre == small) {
                    small = cur;
                    pre = pre.next;
                    cur = cur.next;
                } else {
                    pre.next = cur.next;
                    cur.next = small.next;
                    small.next = cur;
                    small = cur;
                    cur = pre.next;
                }
            } else {
                pre = pre.next;
                cur = cur.next;
            }
        }
        return dummy.next;
    }
}