数据结构-链表中倒数第K个节点

题目：输入一个链表，输出该链表中倒数第K个节点。为了符合大多数人的习惯，本题从1开始计数，即链表的尾节点是倒数第1个节点。例如一个链表有6个节点，从头节点开始他们的值依次为1，2，3，4，5，6.这个链表的倒数第三个节点是值为4的节点。

分析：不能多次遍历链表，只能一次遍历。

/*
剑指offer面试题15
*/
#include <iostream>

using namespace std;

struct ListNode{
    ListNode* Next;
    int data;
};


/*
把题目看错，看成删除了。但是删除功能可以实现，
同样存在和题目一样的问题，存在多次遍历问题。
为了代码质量更好要求只遍历一次。
*/
ListNode* DeleteKnode(ListNode** head,int k){
    if(*head == NULL || k <= 0){
        return NULL;
    }

    ListNode* p = *head;
    ListNode* q = new ListNode;
    int count = 0;
    int num = 0;

    while(p != NULL){
        p = p->Next;
        count++;
    }
    //加入k大于链表长度，非法
    if(k > count){
        return NULL;
    }
    //删除尾节点
    if(k == 1){
        p = *head;
        int i=1;
        while(i<(count-1)){
            p = p->Next;
            i++;
        }
        q = p->Next;
        delete q;
        q = NULL;
        p->Next = NULL;
        return *head;
    }
    //删除头节点
    if(k == count){
        p = *head;
        q = p->Next;
        delete p;
        p = NULL;
        return q;
    }

    else{
        p = *head;
        while(p != NULL && (count-k-1) > num){
            num++;
            p = p->Next;
        }
        q = p->Next;
        p->Next = q->Next;
        delete q;
        q = NULL;
    }

    return p;
}

void Print(ListNode** head,int k){
    if(*head == NULL || k <= 0){
        return;
    }

    ListNode* first = *head;
    int length = 0;
    for(int i=0;i<k-1;++i){
        if(first->Next != NULL){    //这是关键步骤！！
            first = first->Next;
            length++;
        }
        else{
            return;
        }
    }

    ListNode* second = *head;
    while(first->Next != NULL){
        first = first->Next;
        length++;
        second = second->Next;
    }

    cout << second->data << endl;

}

int main()
{
    ListNode* head = new ListNode;
    ListNode* One = new ListNode;
    ListNode* Two = new ListNode;
    ListNode* tail = new ListNode;

    head->data = 0;
    head->Next = One;
    One->data = 1;
    One->Next = Two;
    Two->data = 2;
    Two->Next = tail;
    tail->data = 3;
    tail->Next = NULL;

    int k;

    cin >> k;

/*
    ListNode* p = DeleteKnode(&head,k);

    while(p != NULL){
        cout << p->data << " ";
        p = p->Next;
    }
*/

    Print(&head,k);

    return 0;
}