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->null and x = 3,

return 1->2->2->4->3->5->null.

 

SOLUTION:

其实就是快速排序嘛，比较简单，按照题意实现就对了，开两个头节点，一个放小于x的，一个放大于x的，再拼起来。

注意，注意，注意，链表的最后一定要指向null，就是右边的链表尾部一定要指向null！！！

代码：

/** * Definition for ListNode. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int val) { *         this.val = val; *         this.next = null; *     } * } */ public class Solution {    /**     * @param head: The first node of linked list.     * @param x: an integer     * @return: a ListNode      */    public ListNode partition(ListNode head, int x) {        if (head == null){            return head;        }        ListNode dummyLeft = new ListNode(0);        ListNode dummyRight = new ListNode(0);        ListNode left = dummyLeft;        ListNode right = dummyRight;        while (head != null){            if (head.val < x){                left.next = head;                left = left.next;            } else {                right.next = head;                right = right.next;            }            head = head.next;        }        left.next = dummyRight.next;        right.next = null;        return dummyLeft.next;    }}