2021-03-13：手写代码：单链表快排。
福大大 答案2021-03-13：
根据链表的表头三分。比表头小的元素放左边，比表头大的元素放右边，等于表头的元素放中间。然后递归左边和递归右边。最后合并左、中、右。
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, " ")
cur = cur.Next
}
fmt.Println()
head = sortList(head)
cur = head
for cur != nil {
fmt.Print(cur.Val, " ")
cur = cur.Next
}
fmt.Println()
}
//Definition for singly-linked list.
type ListNode struct {
Val int
Next *ListNode
}
func sortList(head *ListNode) *ListNode {
ret, _ := process(head)
return ret
}
func process(head *ListNode) (*ListNode, *ListNode) {
left, leftend, mid, midend, right, rightend := partition(head)
if left != nil {
left, leftend = process(left)
}
if right != nil {
right, rightend = process(right)
}
return merge(left, leftend, mid, midend, right, rightend)
}
func partition(head *ListNode) (*ListNode, *ListNode, *ListNode, *ListNode, *ListNode, *ListNode) {
left := &ListNode{} //虚拟节点
leftend := left
mid := head
midend := mid
right := &ListNode{} //虚拟节点
rightend := right
cur := head.Next
for cur != nil {
if cur.Val < mid.Val {
leftend.Next = cur
leftend = leftend.Next
} else if cur.Val == mid.Val {
midend.Next = cur
midend = midend.Next
} else {
rightend.Next = cur
rightend = rightend.Next
}
cur = cur.Next
}
leftend.Next = nil
midend.Next = nil
rightend.Next = nil
left = left.Next
if left == nil {
leftend = nil
}
right = right.Next
if right == nil {
rightend = nil
}
return left, leftend, mid, midend, right, rightend
}
func merge(left, leftend, mid, midend, right, rightend *ListNode) (*ListNode, *ListNode) {
head := &ListNode{}
headend := head
if left != nil {
headend.Next = left
headend = leftend
}
headend.Next = mid
headend = midend
if right != nil {
headend.Next = right
headend = rightend
}
head = head.Next
if head == nil {
headend = nil
}
return head, headend
}
执行结果如下：
