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：
