2021-03-14：手寫代碼：單鏈表冒泡排序。

2021-03-14：手寫代碼：單鏈表冒泡排序。

福大大 答案2021-03-14：

遍歷鏈表，算出元素個數，假設爲N。須要嵌套循環，外循環N-1輪，每輪循環相鄰交換N-1次。

代碼用golang編寫，代碼以下：

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
    }

    //鏈表節點的個數
    count := 0

    //對鏈表節點計數
    temp := head.Next
    for temp != nil {
        count++
        temp = temp.Next
    }

    //有換頭的可能，因此新增一個虛擬頭節點
    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 {
                //相鄰交換
                pre.Next, cur.Next, cur.Next.Next, cur = cur.Next, cur.Next.Next, cur, cur.Next
            }
            pre = pre.Next
            cur = cur.Next
        }
    }

    //虛擬頭節點的Next指針就是須要返回的節點
    return preHead.Next
}

執行結果以下：

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力扣148. 排序鏈表
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