题解 | #输出单向链表中倒数第k个结点#

const rl = require("readline").createInterface({ input: process.stdin });
var iter = rl[Symbol.asyncIterator]();
const readline = async () => (await iter.next()).value;
// 方法一：不讲武德版本
void (async function () {
    while ((n = parseInt(await readline()))) {
        const list = (await readline()).split(" ").map(Number); //链表的值
        const k = parseInt(await readline()); //k，即倒数第k个的k
        console.log(list[n - k]);
    }
})();

// 方法二：按题意来做
void (async function () {
    while ((n = parseInt(await readline()))) {
        // 输入
        const list = (await readline()).split(" ").map(Number); //链表的值
        let k = parseInt(await readline()); //k，即倒数第k个的k

        // 构造链表
        const ListNode = (value,next = null) => ({value,next});
        const head = ListNode(list.shift());
        let node = head;
        while(list.length){
            node.next = ListNode(list.shift());
            node = node.next;
        }

        // 求倒数第k个节点的值,快慢指针
        let slow = head,fast = head;
        while(k--) fast = fast.next;
        while(fast){
            slow = slow.next;
            fast = fast.next;
        }
        console.log(slow.value);        
    }
})();