《程序员代码面试指南》第二章 链表问题 将单链表按某值划分为左边小，中间相等，右边大的链表

题目

例如 链表 1-2-3-6-7-8-5-4-9-10 ，按 5分为左边小，中间相等，右边大的链表，1-2-3-4-5-6-7-8-9-10

java代码

/**
 * @Description:将单链表按某值划分为左边小，中间相等，右边大的链表
 * @Author: lizhouwei
 * @CreateDate: 2018/4/6 21:26
 * @Modify by:
 * @ModifyDate:
*/
public class Chapter2_8 {
    public Node partition(Node head, int pivot) {
        if (head == null) {
            return null;
        }
        Node sH = null;//小的链表的头节点
        Node sT = null;//小的链表的尾节点
        Node eH = null;//相等的链表的头节点
        Node eT = null;//相等的链表的头节点
        Node bH = null;//大的链表的头节点
        Node bT = null;//大的链表的尾节点
        Node next = null;
        while (head != null) {
            next = head.next;
            head.next = null;
            if (head.vlaue < pivot) {
                if (sH == null) {
                    sH = head;
                    sT = head;
                } else {
                    sT.next = head;
                    sT = head;
                }
            } else if (head.vlaue == pivot) {
                if (eH == null) {
                    eH = head;
                    eT = head;
                } else {
                    eT.next = head;
                    eT = head;
                }
            } else {
                if (bH == null) {
                    bH = head;
                    bT = head;
                } else {
                    bT.next = head;
                    bT = head;
                }
            }
            head = next;
        }
        //重新连接
        if (sT != null) {
            sT.next = eH;
            eT = eT == null ? sT : eT;
        }
        if (eT != null) {
            eT.next = bH;
        }
        return sH != null ? sH : eH != null ? eH : bH;
    }

    //测试
    public static void main(String[] args) {
        Chapter2_8 chapter = new Chapter2_8();
        Link link = new Link();
        //构造两个链表
        for (int i = 10; i > 0; i--) {
            link.add(i);
            link.add(i + 10);
        }
        Link.printLink(link.head);
        Node head = chapter.partition(link.head, 10);
        Link.printLink(head);
    }
}