Leetcode Partition List 分割链表

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：

对于值大于等于x的节点，直接插到链表。对于值小于x的节点，需要找到第一个大于等于x的节点，插到它的前面。

Java代码实现：

/**
* 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) {
if(head==null || head.next==null)
return head;

ListNode dummy = new ListNode(0);
ListNode node = dummy;
node.next = null;

while(head!=null)
{
if(head.val>=x)
{
while(node.next!=null)
{
node = node.next;
}
node.next = new ListNode(head.val);
node.next.next = null;
}
else
{
while(node.next!=null && node.next.val < x)
{
node = node.next;
}
ListNode temp = node.next;
node.next = new ListNode(head.val);
node.next.next = temp;
}
node = dummy;
head = head.next;
}

return dummy.next;
}
}