归并排序单链表实现

Posted 2020-10-08 Michael2397

/**
 * Project Name:Algorithm
 * File Name:SingleListMergeSort.java
 * Package Name:
 * Date:2017年9月22日上午10:08:46
 * Copyright (c) 2017, [e-mail] All Rights Reserved.
 *
 */

/**
 * ClassName:SingleListMergeSort 
 * Function: 链表实现归并排序
 * Reason:    (1)定义链表结构 
 *             (2)链表的关键在于递归的时候中间位置的确定，方法是：用两个指针slow，fast 遍历链表，slow走一步而fast走两步；当fast走完的时候slow走到链表的一半！
 *             (3)时间复杂度为O(nlgn），空间复杂度为O(1)　　 
 * Date:     2017年9月22日 上午10:08:46 
 * @author   michael
 * @version  
 * @since    JDK 1.7
 * @see      
 */
public class MergeSortSingle {
    //先找到中间点，然后分割
    public ListNodes getMiddle(ListNodes head){
        if(head == null){
            return head;
        }
        //定义快慢指针
        ListNodes slow, fast;
        slow=fast=head;
        while(fast.next!=null&&fast.next.next!=null){
            slow = slow.next;
            fast = fast.next.next;
        }
        return slow;
    }
    public ListNodes merge(ListNodes a, ListNodes b){
        //dummyHead是一个新的链表
        //将dummyHead的头节点赋值给curr，curr往后遍历，而dummyHead还是指向头节点
        ListNodes dummyHead,curr;
        dummyHead = new ListNodes(0);
        curr = dummyHead;
        while (a!=null&&b!=null) {
            if(a.val<b.val){
                curr.next=a;
                a = a.next;
            }else {
                curr.next = b;
                b = b.next;
            }
            //curr往后移动
            curr = curr.next;
        }
        //如果a遍历完，则将b剩下的部分赋值给curr
        curr.next = (a!=null)?a:b;
        return dummyHead.next;
    }
    public ListNodes merge_sort(ListNodes head){
        if(head==null||head.next==null){
            return head;
        }
        //取链表中间节点，作为前部分的最后一个节点
        ListNodes middle = getMiddle(head);
        //中间节点的下一个节点，作为后部分的后一个节点
        ListNodes halfHead = middle.next;
        //分：在中间节点处断开
        middle.next=null;
        //合：
        return merge(merge_sort(head), merge_sort(halfHead));
    }

    public static void main(String[] args) {
        ListNodes head = new ListNodes(0);
        ListNodes l1 = new ListNodes(2);
        ListNodes l2 = new ListNodes(5);
        ListNodes l3 = new ListNodes(3);
        ListNodes l4 = new ListNodes(8);
        ListNodes l5 = new ListNodes(5);
        ListNodes l6 = new ListNodes(2);
        ListNodes l7 = new ListNodes(1);

        head.next = l1;
        l1.next = l2;
        l2.next = l3;
        l3.next = l4;
        l4.next = l5;
        l5.next = l6;
        l6.next = l7;
        l7.next = null;

        ListNodes p = head;
        while (p.next != null) {
            System.out.print(p.val);
            p = p.next;
        }
        System.out.print(p.val);
        System.out.println();
        new MergeSortSingle().merge_sort(head);
        
        p = head;
        while(p!=null){
            System.out.print(p.val);
            p = p.next;
        }
    }
}

class ListNodes{
    //节点的值
    int val;
    //连接到下一个节点
    ListNodes next;
    
    public ListNodes(int x) {
        val = x;
        next = null;
    }
}