2021-03-15：手写代码：单链表选择排序。

2021-03-15：手写代码：单链表选择排序。

福大大 答案2021-03-15：

遍历链表，找出最小元素，链表里删除最小元素，最小元素放在需要返回的链表里。

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 = SelectSort(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
}

//选择排序
func SelectSort(head *ListNode) *ListNode {
    if head == nil || head.Next == nil {
        return head
    }

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

    //选择
    var pre, cur, preSel, sel *ListNode
    for preHead.Next != nil {
        pre, cur = preHead, preHead.Next

        //默认选中第1个节点
        preSel, sel = pre, cur

        //选最小的，从第2个节点开始
        pre, cur = cur, cur.Next
        for cur != nil {
            if cur.Val < sel.Val {
                preSel, sel = pre, cur
            }
            pre, cur = cur, cur.Next
        }

        //选中的节点放在答案里
        preAnsEnd.Next = sel

        //原链表删除选中的节点
        preSel.Next = sel.Next

        //尾指针指向Next
        preAnsEnd = preAnsEnd.Next
    }

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

执行结果如下：

力扣148. 排序链表
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