题解 | 输出单向链表中倒数第k个结点

#include <iostream>

using namespace std;

struct ListNode{
    int val;
    struct ListNode* m_pnext;
};
int main(){
    int n; // 链表节点个数
    int k; 
    while(cin>>n){
        // 构造头节点
        ListNode* dummy = new ListNode{0,nullptr};
        ListNode* cur = dummy;
        for(int i = 0; i < n; ++i){
            int value;
            cin>>value;
            ListNode* newNode = new ListNode{value,nullptr};
            cur->m_pnext = newNode;
            cur = newNode;
        }
        // 输出第k个节点的值
        cin>>k;
        int count = 0;
        cur = dummy;
        while(count < k){
            cur = cur->m_pnext;
            count++;
        }
        // cur指向正向的第k个节点
        ListNode* pCur = dummy;
        while(cur!=nullptr){
            pCur = pCur->m_pnext;
            cur = cur->m_pnext;
        }
        cout<<pCur->val<<endl;
    }
}