2-3 Reversing Linked List

02-线性结构3 Reversing Linked List （25 分）

Given a constant K and a singly linked list L, you are supposed to reverse the links of every K elements on L. For example, given L being 1→2→3→4→5→6, if K=3, then you must output 3→2→1→6→5→4; if K=4, you must output 4→3→2→1→5→6.

Input Specification:

Each input file contains one test case. For each case, the first line contains the address of the first node, a positive N (≤10^​5​​ ) which is the total number of nodes, and a positive K (≤N) which is the length of the sublist to be reversed. The address of a node is a 5-digit nonnegative integer, and NULL is represented by -1.

Then N lines follow, each describes a node in the format:

Address Data Next

where Address is the position of the node, Data is an integer, and Next is the position of the next node.

Output Specification:

For each case, output the resulting ordered linked list. Each node occupies a line, and is printed in the same format as in the input.

Sample Input:

00100 6 4
00000 4 99999
00100 1 12309
68237 6 -1
33218 3 00000
99999 5 68237
12309 2 33218

Sample Output:

00000 4 33218
33218 3 12309
12309 2 00100
00100 1 99999
99999 5 68237
68237 6 -1

理解题意：

• 输入的第一行很重要：三个数分别代表：首元素的地址，元素总个数N，前K个元素做逆序
  接下来就是N行数据输入，三个数分别代表，该元素地址，该元素，该元素指向的下个节点的地址
• 输出就很容易理解了

我的答案：

// C++版本，更简单一些

#include
#include
#include    ///使用到reverse 翻转函数
using namespace std;

#define MAXSIZE 1000010   ///最大为五位数的地址

struct node    ///使用顺序表存储data和下一地址next
{
    int data;
    int next;
}node[MAXSIZE];

int List[MAXSIZE];   ///存储可以连接上的顺序表
int main()
{
    int First, n, k;
    cin>>First>>n>>k;   ///输入头地址 和 n，k；
    int Address,Data,Next;
    for(int i=0;i>Address>>Data>>Next;
        node[Address].data=Data;
        node[Address].next=Next;
    }

    int j=0;  ///j用来存储能够首尾相连的节点数
    int p=First;   ///p指示当前结点
    while(p!=-1)
    {
        List[j++]=p;
        p=node[p].next;
    }
    int i=0;
    while(i+k<=j)   ///每k个节点做一次翻转
    {
        reverse(&List[i],&List[i+k]);
        i=i+k;
    }
    for(i=0;i

c版本：

// C 实现版本， 与老师讲的一致

#include 
#include 
#define MaxSize 100000
typedef int Ptr;
//结构数组模拟链表
struct node{
    int key;
    Ptr next;
}list[MaxSize];

//逆转某一段含K个结点的链表
Ptr Reverse ( Ptr head, int K ) {
    Ptr New, Old, Tmp; //New指向逆转链表表头，Old指向未逆转表头，Tmp记录Old后一个结点
    int cnt = 1;
    New = head;
    Old = list[New].next;
    while ( cnt < K ) {
        Tmp = list[Old].next; //记录Old下一个结点，防止逆转指针后丢失
        list[Old].next = New; //逆转指针
        New = Old; Old = Tmp; //New，Old向后转移一个
        cnt++;
    }
    list[head].next = Old; //原来的第一个结点逆转后变为最后一个结点，并连接剩下未逆转的元素
    return New;
}

//判断是否需要逆转
int NeedReverse ( Ptr head , int K) {
    int i;
    for( i = 1; list[head].next != -1; head = list[head].next ) {
        i++;
        if ( i == K ) return 1; //还有K个或以上结点，需要逆转
    }
    return 0; //不足K个结点，不需要逆转
}

//逆转整个链表
Ptr ReversingLinkedList( Ptr head , int K ) {
    Ptr UnreversedHead = head;  //未逆转的链表的第一个结点
    Ptr ListHead; //整个表的第一个结点
    Ptr TempTail; //临时表尾,用来连接下一段逆转的链表

    if ( NeedReverse( UnreversedHead, K ) ) { //第一次先判断是否需要逆转
        ListHead = Reverse( UnreversedHead, K ); //记住逆转后的整个链表的第一个结点
        TempTail = UnreversedHead; //记录此逆转链表的表尾
        UnreversedHead = list[TempTail].next; //记录未逆转链表的表头
    }
    else //链表结点个数小于K，无需逆转
        return head;

    while ( NeedReverse( UnreversedHead, K ) ) {
        list[TempTail].next = Reverse( UnreversedHead, K ); //上一个逆转链表的表尾与这个逆转链表的表头连接
        TempTail = UnreversedHead;
        UnreversedHead = list[TempTail].next;
    }
    return ListHead;
}

//输出链表
void PrintLinkedList ( Ptr head ) {
    Ptr temp = head;
    for(; list[temp].next !=  -1;  temp = list[temp].next)
        printf("%05d %d %05d\n", temp, list[temp].key, list[temp].next);
    printf("%05d %d %d\n", temp, list[temp].key, list[temp].next);
}

int main(){
    Ptr ad, head;
    int N, K;
    scanf("%d %d %d", &head, &N, &K);
    for(int i = 0; i < N; i++){
        scanf("%d", &ad);
        scanf("%d %d", &list[ad].key, &list[ad].next);
    }
    PrintLinkedList( ReversingLinkedList( head, K ) );
    return 0;
}

答案解析

明天再补上！！