题解 | #输出单向链表中倒数第k个结点#

#include <bits/stdc++.h>
using namespace std;
struct ListNode{
    int m_nKey;
    ListNode* m_pNext;
    ListNode():m_nKey(0),m_pNext(NULL){}
    ListNode(int x):m_nKey(x),m_pNext(NULL){}
};
int main(){
    int n,k,val,i;
    while(cin>>n){
         ListNode* head=new ListNode();//带有头结点，注意头结点没有存储数字
        ListNode* curr=head;
        for(i=0;i<n;i++){//正向建立链表
            cin>>val;
            ListNode* temp=new ListNode(val);
            head->m_pNext=temp;
            head=head->m_pNext;
        }
        cin>>k;
        ListNode *fast=curr,*slow=curr;//快慢指针
        for(i=0;i<k;i++)//先让快指针移动k步
            fast=fast->m_pNext;
        while(fast!=NULL){//当快指针到达尾部时，此时慢指针指向的就是倒数第k个节点
            slow=slow->m_pNext;
            fast=fast->m_pNext;
        }
        cout<<slow->m_nKey<<"\n";;
    }
    return 0;
}