寻找单链表的倒数第N个节点

来源：互联网发布：自动化编程语言有哪些编辑：程序博客网时间：2024/05/16 08:04

老婆今早面试苹果的radio协议测试岗位（非开发），给了这个题目，当时她是两次遍历（一次找出总节点数M，一次找出第M-N节点），我也想不出更高明的办法

但总觉得伴随指针能做到只遍历一遍，想了想，得解：

#include <iostream>using namespace std;struct Node{    Node(int d, Node *n):data(d), next(n){}    int data;    struct Node *next;};struct List{    List(const char *c):listHead(0, NULL)    {        Node *p = &listHead;        for (; *c; c++)        {            p->next = new Node(*c, NULL);            p = p->next;            listHead.data++;// node count        }    }    Node *FindNthNodeFromBottom(int N)    {        Node *p = listHead.next;        Node *q = p;        int step = 0;        while (p != NULL)        {            if (step >= N)            {                q = q->next;//when q left N steps behind p, then q start move            }else{                step++;            }            p = p->next;        }        return q;    }    void printList()    {        Node *p = listHead.next;        while (p)        {            cout << (char) (p->data) << ' ';            p = p-> next;        }        cout << endl;    }    Node listHead;};int main(){    List l("wanghaipeng");    l.printList();    int N = 4;    Node *p = l.FindNthNodeFromBottom(N);    cout << "the " << N <<"th node from bottom of list is : "         << (char)(p->data) << endl;    return 0;}

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