2021-03-14：手写代码：单链表冒泡排序。

2021-03-14：手写代码：单链表冒泡排序。

福大大 答案2021-03-14：

遍历链表，算出元素个数，假设为N。需要嵌套循环，外循环N-1轮，每轮循环相邻交换N-1次。

代码用golang编写，代码如下：

package main

import "fmt"

func main() {
    //head := &ListNode{Val: 4}
    //head.Next = &ListNode{Val: 2}
    //head.Next.Next = &ListNode{Val: 1}
    //head.Next.Next.Next = &ListNode{Val: 3}

    head := &ListNode{Val: -1}
    head.Next = &ListNode{Val: 5}
    head.Next.Next = &ListNode{Val: 3}
    head.Next.Next.Next = &ListNode{Val: 4}
    head.Next.Next.Next.Next = &ListNode{Val: 0}

    cur := head
    for cur != nil {
        fmt.Print(cur.Val, "\t")
        cur = cur.Next
    }
    fmt.Println()

    head = BubbleSort(head)

    cur = head
    for cur != nil {
        fmt.Print(cur.Val, "\t")
        cur = cur.Next
    }
    fmt.Println()
}

//Definition for singly-linked list.
type ListNode struct {
    Val  int
    Next *ListNode
}

//冒泡排序
func BubbleSort(head *ListNode) *ListNode {
    if head == nil || head.Next == nil {
        return head
    }

    //链表节点的个数
    count := 0

    //对链表节点计数
    temp := head.Next
    for temp != nil {
        count++
        temp = temp.Next
    }

    //有换头的可能，所以新增一个虚拟头节点
    preHead := &ListNode{Next: head}

    //冒泡
    var pre, cur *ListNode
    for i := 0; i < count; i++ {
        pre = preHead
        cur = preHead.Next
        for j := 0; j < count-i; j++ {
            if cur.Val > cur.Next.Val {
                //相邻交换
                pre.Next, cur.Next, cur.Next.Next, cur = cur.Next, cur.Next.Next, cur, cur.Next
            }
            pre = pre.Next
            cur = cur.Next
        }
    }

    //虚拟头节点的Next指针就是需要返回的节点
    return preHead.Next
}

执行结果如下：

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力扣148. 排序链表
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