题解 | #输出单向链表中倒数第k个结点#

package main


import (
	"bufio"
	"fmt"
	"os"
	"strconv"
    "strings"
)

func main() {
	type ListNode struct {
		m_nKey  int
		m_pNext *ListNode
	}

	in := bufio.NewScanner(os.Stdin)
	for in.Scan() {
		n, _ := strconv.Atoi(in.Text())

		list := new(ListNode)
		head := list

		in.Scan()
		arr := strings.Fields(in.Text())
		for i := 0; i < n; i++ {
			key, _ := strconv.Atoi(arr[i])
			node := &ListNode{
				m_nKey: key,
			}
			head.m_pNext = node
			head = node
		}

		in.Scan()
		k, _ := strconv.Atoi(in.Text())
		// slow == 倒数 k <==> fast n - k + 1
		fast, slow := list, list
		for i := 0; i < k; i++ {
			fast = fast.m_pNext
		}

		for fast != nil {
			fast = fast.m_pNext
			slow = slow.m_pNext
		}

		if slow == nil {
			fmt.Println(slow)

			return
		}

		fmt.Println(slow.m_nKey)
	}
}