2021-03-16：手写代码：单链表归并排序。

福大大答案2021-03-16：

获取链表中点，然后按中点分成两个链表。递归两个链表。合并两个链表。

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}

    printlnLinkNodeList(head)

    head = MergeSort(head)

    printlnLinkNodeList(head)
}

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

//链表打印
func printlnLinkNodeList(head *ListNode) {
    cur := head
    for cur != nil {
        fmt.Print(cur.Val, "\t")
        cur = cur.Next
    }
    fmt.Println()
}

//归并排序
func MergeSort(head *ListNode) *ListNode {
    if head == nil || head.Next == nil {
        return head
    }
    ret := process(head)
    return ret
}

//递归，head不可能为空
func process(head *ListNode) *ListNode {
    //如果只有1个节点，直接返回
    if head.Next == nil {
        return head
    }

    //获取中点
    mid := getMid(head)
    midNext := mid.Next

    //按中点拆分链表
    mid.Next = nil

    //递归
    left := process(head)
    right := process(midNext)

    //合并
    return merge(left, right)

}

//找中点，head一定不为空
func getMid(head *ListNode) *ListNode {
    fast := head
    slow := head
    for fast.Next != nil && fast.Next.Next != nil {
        fast = fast.Next.Next
        slow = slow.Next
    }
    return slow
}

//合并，left和right一定都不为空
func merge(left *ListNode, right *ListNode) *ListNode {
    preAns := &ListNode{}
    preAnsEnd := preAns

    for left != nil && right != nil {
        if left.Val <= right.Val {
            preAnsEnd.Next = left
            left = left.Next
        } else {
            preAnsEnd.Next = right
            right = right.Next
        }
        preAnsEnd = preAnsEnd.Next
    }

    if left == nil {
        preAnsEnd.Next = right
    } else {
        preAnsEnd.Next = left
    }

    return preAns.Next
}