LeetCode

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.

思路：

分成两个链表，然后链接

package list;

public class PartitionList {

    public ListNode partition(ListNode head, int x) {
        ListNode leftNode = new ListNode(0);
        ListNode rightNode = new ListNode(0);
        ListNode p = leftNode;
        ListNode q = rightNode;
        while (head != null) {
            if (head.val < x) {
                leftNode.next = head;
                leftNode = leftNode.next;
            } else {
                rightNode.next = head;
                rightNode = rightNode.next;
            }
            head = head.next;
        }
        rightNode.next = null;
        leftNode.next = q.next;
        return p.next;
    }
    
    public static void main(String[] args) {
        // TODO Auto-generated method stub
        ListNode a1 = new ListNode(1);
        ListNode a2 = new ListNode(4);
        ListNode a3 = new ListNode(3);
        ListNode a4 = new ListNode(2);
        ListNode a5 = new ListNode(5);
        ListNode a6 = new ListNode(2);
        a1.next = a2;
        a2.next = a3;
        a3.next = a4;
        a4.next = a5;
        a5.next = a6;
        a6.next = null;
        PartitionList p = new PartitionList();
        ListNode head = p.partition(a1, 3);
        while (head != null) {
            System.out.println(head.val);
            head = head.next;
        }
    }

}