单链表的归并排序和插入排序-蒲公英云

单链表的归并排序和插入排序

男娘i 2022-06-13 04:21 319阅读 0赞

由于最近在学习数据结构和算法，在牛客网的在线编程题上遇到了对链表的相关排序操作,发现自己对链表这块还是理解不够深入，以前做过对数组进行排序，但链表的操作要比数组复杂一些，毕竟不能进行随机访问了，单链表得从头开始找。
下面是我通过在线编程（OJ）代码并附有注释。

1.单链表的归并排序

1.题目

Sort a linked list in O(n log n) time using constant space complexity.
对一个链表进行排序，要求时间复杂度为O(n log n)

2.解题思路

题目要求时间复杂度为nlog(n),所以选择了归并排序。归并排序最坏时间复杂度、平均时间复杂度、最快时间复杂度都为nlog(n)。它是一种稳定的排序算法。
链接：思路来源
来源：牛客网
这里写图片描述

2.1 拆分链表,找到中间节点

用快慢指针的方式，快指针移动两位，慢指针移动一位。

/** 查找中间结点 */
public ListNode findMidNode(ListNode head){
    ListNode p= head;
    ListNode q=head.next;
    while (q != null && q.next != null) {
        p = p.next;
        q = q.next.next;
    }
    return p;
}

2.2分治，分别处理每个链表

ListNode midNode =findMidNode(head);
    ListNode record = midNode.next;
    midNode.next = null; //断开两个链表的连接
    ListNode left = MySort(head, midNode);
    ListNode right = MySort(record, end);

2.3合并

/** 合并左右链表 */
private ListNode merge(ListNode left, ListNode right) {
    if(left==null)return right;
    if(right==null)return left;
    ListNode newhead = null;
    newhead=new ListNode(0);
    ListNode temp = newhead;
    while (left != null && right != null) {
        if (left.val > right.val) {
            temp.next= right; 
            right= right.next;
        } else {
            temp.next= left;
            left = left.next;
        }   
        temp=temp.next;
    }
    temp.next=(left==null)?right:left;
    return newhead.next;
}

3. 完整代码

/** * Definition for singly-linked list. * class ListNode { * int val; ListNode next; * ListNode(int x) { * val = x; next = null; * } *} //排序算法选择归并排序 */
public class Solution { 
    private ListNode indexNode =new ListNode(Integer.MAX_VALUE); 
    public ListNode sortList(ListNode head) {
        if (head == null) {
            return head;
        }
        return MySort(head, null);
    }
    private ListNode MySort(ListNode head, ListNode end) {
        if(head==end){
            return head;
        }
        ListNode midNode =findMidNode(head);
        ListNode record = midNode.next;
        midNode.next = null;
        ListNode left = MySort(head, midNode);
        ListNode right = MySort(record, end);
        return merge(left, right);
    }
     /** 查找中间结点 */
    public ListNode findMidNode(ListNode head){
        ListNode p= head;
        ListNode q=head.next;
        while (q != null && q.next != null) {
            p = p.next;
            q = q.next.next;
        }
        return p;
    }
    /** 合并左右链表 */
    private ListNode merge(ListNode left, ListNode right) {
        if(left==null)return right;
        if(right==null)return left;
        ListNode newhead = null;
        newhead=new ListNode(0);
        ListNode temp = newhead;
        while (left != null && right != null) {
            if (left.val > right.val) {
                temp.next= right; 
                right= right.next;
            } else {
                temp.next= left;
                left = left.next;
            }   
            temp=temp.next;
        }
        temp.next=(left==null)?right:left;
        return newhead.next;
    }
    public static void main(String[] args) {
        ListNode list = new ListNode(4);
        ListNode head = list;
        head.next = new ListNode(2);
        head.next.next = new ListNode(1);
        head.next.next.next = new ListNode(3);
        Solution solution = new Solution();
        ListNode result = solution.sortList(head);
        while (result != null) {
            System.out.print(result.val + "\t");
            result = result.next;
        }
    }
    public void display(ListNode result) {
        while (result != null) {
            System.out.print(result.val + "\t");
            result = result.next;
        }
        System.out.println();
    }
}
class ListNode {
    int val;
    ListNode next;
    ListNode(int x) {
        val = x;
        next = null;
    }
}

2.单链表的直接插入排序

1.题目

Sort a linked list using insertion sort.

2.注意事项

每次的插入操作时，需要保留下一结点。
然后用伪首结点该结点值为最小值，将他们串联起来。

完整代码

public class Solution {
    public ListNode insertionSortList(ListNode head) {
        if (head == null || head.next == null) // 链表为空或者仅一个结点
            return head;
        ListNode saveNode = new ListNode(Integer.MIN_VALUE);
        // ListNode current =head;
        ListNode previous = saveNode;
        while (head != null) {
            ListNode next = head.next;
            previous = saveNode; // 每次都从新链表的头结点进行遍历
            while (previous.next != null && previous.next.val < head.val) {
                previous = previous.next;
            }
            if (previous == saveNode) { // 链表为空时
                head.next = previous.next;
                previous.next=head;
            } else if(previous.next==null){  //尾部结点时
                head.next = previous.next;
                previous.next=head;
            }
            else {  //中间插入
                ListNode nextnext=previous.next;
                head.next=nextnext;
                previous.next=head;
            }
            head = next;
        }
        return saveNode.next;
    }
    public void display(ListNode node) {
        while (node != null) {
            System.out.print(node.val + "\t");
            node = node.next;
        }
    }
}
class ListNode {
    int val;
    ListNode next;
    ListNode(int x) {
        val = x;
        next = null;
    }
}