給定一個單連結清單 L,我們将每 K 個結點看成一個區塊(連結清單最後若不足 K 個結點,也看成一個區塊),請編寫程式将 L 中所有區塊的連結反轉。例如:給定 L 為 1→2→3→4→5→6→7→8,K 為 3,則輸出應該為 7→8→4→5→6→1→2→3。
輸入格式:
每個輸入包含 1 個測試用例。每個測試用例第 1 行給出第 1 個結點的位址、結點總個數正整數 N (≤105)、以及正整數 K (≤N),即區塊的大小。結點的位址是 5 位非負整數,NULL 位址用 −1 表示。
接下來有 N 行,每行格式為:
Address Data Next
輸出格式:
輸入樣例:
00100 8 3
71120 7 88666
00000 4 99999
00100 1 12309
68237 6 71120
33218 3 00000
99999 5 68237
88666 8 -1
12309 2 33218
輸出樣例:
71120 7 88666
88666 8 00000
00000 4 99999
99999 5 68237
68237 6 00100
00100 1 12309
12309 2 33218
33218 3 -1
代碼實作:
import java.io.*;
import java.util.ArrayList;
/**
* @author yx
* @date 2022-07-29 1:32
*/
public class Main {
static PrintWriter out=new PrintWriter(System.out);
static BufferedReader ins=new BufferedReader(new InputStreamReader(System.in));
static StreamTokenizer in=new StreamTokenizer(ins);
static class Node{
int Address;
int Data;
int Next;
Node(int Address,int Data,int Next){
this.Address=Address;
this.Data=Data;
this.Next=Next;
}
}
public static void main(String[] args) throws IOException {
in.nextToken();
int initAddress=(int) in.nval;
in.nextToken();
int N=(int) in.nval;
in.nextToken();
int K=(int) in.nval;
ArrayList<Node> list=new ArrayList<>();//原始連結清單
ArrayList<Node> list1=new ArrayList<>();//區塊反轉連結清單
Node[] nums=new Node[100000];//初始化數組
for (int i = 0; i < N; i++) {
in.nextToken();
int address=(int) in.nval;
in.nextToken();
int data=(int) in.nval;
in.nextToken();
int next=(int) in.nval;
nums[address]=new Node(address,data,next);
}
//按序串成一個連結清單
while (initAddress!=-1){
list.add(nums[initAddress]);
initAddress=nums[initAddress].Next;
}
// for (int i = 0; i < list.size(); i++) {
// System.out.println(list.get(i).Address+" "+list.get(i).Data+" "+list.get(i).Next);
// }
int temp=list.size()/K;
int yu=list.size()%K;
//區塊反轉
if(yu==0){
for (int i = temp; i >0; i--) {
for (int j = (i-1)*K; j <i*K ; j++) {
list1.add(list.get(j));
}
}
}else {
for (int i = temp*K; i <list.size() ; i++) {
list1.add(list.get(i));
}
for (int i = temp; i >0; i--) {
for (int j = (i-1)*K; j <i*K ; j++) {
list1.add(list.get(j));
}
}
}
//類似于NO1025可以卡bug
for (int i = 0; i < list1.size()-1; i++) {
out.printf("%05d %d %05d\n",list1.get(i).Address,list1.get(i).Data,list1.get(i+1).Address);
}
out.printf("%05d %d -1\n",list1.get(list1.size()-1).Address,list1.get(list1.size()-1).Data);
out.flush();
}
}