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<iostream>
#include<stdio.h>
#include<algorithm> ///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<n;i++)
{
cin>>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<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 实现版本， 与老师讲的一致
#include <stdio.h>
#include <stdlib.h>
#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！！
