02-線性結構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

理解題意：

• 輸入的第一行很重要：三個數(shù)分別代表：首元素的地址，元素總個數(shù)N，前K個元素做逆序
  接下來就是N行數(shù)據(jù)輸入，三個數(shù)分別代表，該元素地址，該元素，該元素指向的下個節(jié)點的地址
• 輸出就很容易理解了

我的答案：

// C++版本，更簡單一些

#include<iostream>
#include<stdio.h>
#include<algorithm>    ///使用到reverse 翻轉函數(shù)
using namespace std;

#define MAXSIZE 1000010   ///最大為五位數(shù)的地址

struct node    ///使用順序表存儲data和下一地址next
{
    int data;
    int next;
}node[MAXSIZE];

int List[MAXSIZE];   ///存儲可以連接上的順序表
int main()
{
    int First, n, k;
    cin>>First>>n>>k;   ///輸入頭地址 和 n，k；
    int Address,Data,Next;
    for(int i=0;i<n;i++)
    {
        cin>>Address>>Data>>Next;
        node[Address].data=Data;
        node[Address].next=Next;
    }

    int j=0;  ///j用來存儲能夠首尾相連的節(jié)點數(shù)
    int p=First;   ///p指示當前結點
    while(p!=-1)
    {
        List[j++]=p;
        p=node[p].next;
    }
    int i=0;
    while(i+k<=j)   ///每k個節(jié)點做一次翻轉
    {
        reverse(&List[i],&List[i+k]);
        i=i+k;
    }
    for(i=0;i<j-1;i++)
        printf("%05d %d %05d\n",List[i],node[List[i]].data,List[i+1]);
    printf("%05d %d -1\n",List[i],node[List[i]].data);
    return 0;
}

c版本：

// C 實現(xiàn)版本， 與老師講的一致

#include <stdio.h>
#include <stdlib.h>
#define MaxSize 100000
typedef int Ptr;
//結構數(shù)組模擬鏈表
struct node{
    int key;
    Ptr next;
}list[MaxSize];

//逆轉某一段含K個結點的鏈表
Ptr Reverse ( Ptr head, int K ) {
    Ptr New, Old, Tmp; //New指向逆轉鏈表表頭，Old指向未逆轉表頭，Tmp記錄Old后一個結點
    int cnt = 1;
    New = head;
    Old = list[New].next;
    while ( cnt < K ) {
        Tmp = list[Old].next; //記錄Old下一個結點，防止逆轉指針后丟失
        list[Old].next = New; //逆轉指針
        New = Old; Old = Tmp; //New，Old向后轉移一個
        cnt++;
    }
    list[head].next = Old; //原來的第一個結點逆轉后變?yōu)樽詈笠粋€結點，并連接剩下未逆轉的元素
    return New;
}

//判斷是否需要逆轉
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; //還有K個或以上結點，需要逆轉
    }
    return 0; //不足K個結點，不需要逆轉
}

//逆轉整個鏈表
Ptr ReversingLinkedList( Ptr head , int K ) {
    Ptr UnreversedHead = head;  //未逆轉的鏈表的第一個結點
    Ptr ListHead; //整個表的第一個結點
    Ptr TempTail; //臨時表尾,用來連接下一段逆轉的鏈表

    if ( NeedReverse( UnreversedHead, K ) ) { //第一次先判斷是否需要逆轉
        ListHead = Reverse( UnreversedHead, K ); //記住逆轉后的整個鏈表的第一個結點
        TempTail = UnreversedHead; //記錄此逆轉鏈表的表尾
        UnreversedHead = list[TempTail].next; //記錄未逆轉鏈表的表頭
    }
    else //鏈表結點個數(shù)小于K，無需逆轉
        return head;

    while ( NeedReverse( UnreversedHead, K ) ) {
        list[TempTail].next = Reverse( UnreversedHead, K ); //上一個逆轉鏈表的表尾與這個逆轉鏈表的表頭連接
        TempTail = UnreversedHead;
        UnreversedHead = list[TempTail].next;
    }
    return ListHead;
}

//輸出鏈表
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

明天再補上??！