[Lintcode] Partition List

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;
    }
}