剑指Offer系列---(2)求链表中的倒数第k个结点

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1.题目描述：
求链表中的倒数第k个结点
2.考虑情况：
1）输入的指针为空；
2）结点总数小于k；
3）输入的k为0。
3.拓展情况：
1）求链表的中间结点；
2）判断一个单向链表是否构成了环形结构。（尝试定义两个指针遍历链表，一个指针遍历的速度要比另外一个指针要快，也就是一个一次走一步，另一个一次走两步或者多步）
4.源代码：

//  Copyright (c) 2015年 skewrain. All rights reserved.using namespace std;#include <iostream>#include <stdio.h>struct ListNode{    int m_nValue;    ListNode *m_pNext;};ListNode *CreateLink(int a[],int k){    ListNode *Head = NULL,*q=NULL;    for (int i=0; i<k; i++) {        ListNode *pNew = new ListNode();        pNew->m_nValue = a[i];        pNew->m_pNext = NULL;                if (Head == NULL) {            Head = pNew;            q=pNew;        }        else        {            q->m_pNext=pNew;            q=q->m_pNext;        }    }    return Head;}//从头到尾打印链表void printLink(ListNode *pHead){    cout<<"链表内容为:";    ListNode *p = pHead;    while (p) {        cout<<p->m_nValue<<"";        p=p->m_pNext;    }    cout<<endl;}//求链表中的倒数第k个结点ListNode *FindKthToTail(ListNode *pListHead,unsigned int k){        if(pListHead == NULL || k ==0)        return NULL;    ListNode *pAhead = pListHead;    ListNode *pBehind = NULL;    for(unsigned int i=0;i<k-1;++i)    {        if(pAhead->m_pNext!=NULL)            pAhead = pAhead->m_pNext;        else        {            return NULL;        }    }    pBehind = pListHead;    while (pAhead->m_pNext!=NULL) {        pAhead = pAhead->m_pNext;        pBehind = pBehind->m_pNext;    }    return pBehind;}//========测试用例========//1.pListHead为空void Test1(){    cout<<"测试用例1"<<endl;        ListNode *ptr = NULL;    printLink(ptr);    ListNode *p=FindKthToTail(NULL, 0);    if(p)        cout<<p->m_nValue<<endl;}//2.长度为n,k>n时void Test2(){    cout<<"测试用例2"<<endl;        int a[]={1,2,3,4,5};    ListNode *ptr = CreateLink(a,5);    printLink(ptr);    ListNode *p = FindKthToTail(ptr, 6);    if(p)        cout<<p->m_nValue<<endl;}//3.k==0void Test3(){    cout<<"测试用例3"<<endl;    int a[]={1,2,3,4,5};    ListNode *ptr = CreateLink(a,5);    printLink(ptr);    ListNode *p = FindKthToTail(ptr, 6);    if(p)        cout<<p->m_nValue<<endl;}//4.k==1void Test4(){    cout<<"测试用例4"<<endl;    int a[]= {1,2,3,4,5};    ListNode *ptr = CreateLink(a,5);    printLink(ptr);    ListNode *p = FindKthToTail(ptr, 1);    if(p)        cout<<p->m_nValue<<endl;}//5.k==1,n==1void Test5(){    cout<<"测试用例5"<<endl;    int a[]={1};    ListNode *ptr = CreateLink(a, 1);    printLink(ptr);    ListNode *p = FindKthToTail(ptr, 1);    if(p)        cout<<p->m_nValue<<endl;}//6.normalvoid Test6(){    cout<<"测试用例6"<<endl;    int a[]={1,2,3,4,5};    ListNode *ptr = CreateLink(a, 5);    printLink(ptr);    ListNode *p = FindKthToTail(ptr, 4);    if (p)        cout<<p->m_nValue<<endl;}int main(int argc, const char * argv[]) {        Test1();    Test2();    Test3();    Test4();    Test5();    Test6();   /*    //手动创建链表    int n,k;    while (scanf("%d %d",&n,&k) != EOF) {        int i,data;        scanf("%d",&data);        ListNode  *p = new ListNode();        if (p) {            exit(EXIT_FAILURE);        }        p->m_nValue = data;        p->m_pNext = NULL;                ListNode *pCur = p;                for (i=0; i<n-1; i++) {            scanf("%d",&data);            ListNode *pNew = new ListNode();            if(pNew == NULL)                exit(EXIT_FAILURE);            pNew->m_nValue = data;            pNew->m_pNext = NULL;            pCur->m_pNext = pNew;            pCur = pCur->m_pNext;        }        ListNode *pFind = FindKthToTail(p, k);        if(pFind == NULL)            printf("NULL");        else            printf("%d\n",pFind->m_nValue);    }*/        return 0;}

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