數據結構-鏈表中倒數第K個節點

題目：輸入一個鏈表，輸出該鏈表中倒數第K個節點。爲了符合大多數人的習慣，本題從1開始計數，即鏈表的尾節點是倒數第1個節點。例如一個鏈表有6個節點，從頭節點開始他們的值依次爲1，2，3，4，5，6.這個鏈表的倒數第三個節點是值爲4的節點。

分析：不能屢次遍歷鏈表，只能一次遍歷。

/*
劍指offer面試題15
*/
#include <iostream>

using namespace std;

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


/*
把題目看錯，當作刪除了。可是刪除功能能夠實現，
一樣存在和題目同樣的問題，存在屢次遍歷問題。
爲了代碼質量更好要求只遍歷一次。
*/
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大於鏈表長度，非法
    if(k > count){
        return NULL;
    }
    //刪除尾節點
    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;
    }
    //刪除頭節點
    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){    //這是關鍵步驟！！
            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;
}