1074 Reversing Linked List (25 分) java 题解

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(≤105) 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

解题思路：

题目大意为给定一个链表，将链表上的每n个元素反转。（部分反转），并输出最终链表。

读取所给链表后，按顺序装入list中，求出需要反转的次数（要用list.size()求，有无效结点），利用双层for处理反转，外层控制次数，内层将每轮的处理元素装入list2中，用list.remove()方便，再将list中的剩余元素装入list2，最后处理结点间关系。

// 处理结点间关系

temp = -1;
       for(int i = list2.size() - 1;i >= 0;i--) {
           list2.get(i).next = temp;
           temp = list2.get(i).adress;;
       }

java代码（超时）：

import java.io.*;
import java.util.*;public class Main {public static void main(String[] args) throws IOException {BufferedReader br = new BufferedReader(new InputStreamReader(System.in));String[] split = br.readLine().split(" ");int adressStart = Integer.parseInt(split[0]);int n = Integer.parseInt(split[1]);int before = Integer.parseInt(split[2]);Node1074 arr[] = new Node1074[100005];for(int i = 0; i < n;i++) {split = br.readLine().split(" ");int adress = Integer.parseInt(split[0]);int data = Integer.parseInt(split[1]);int next = Integer.parseInt(split[2]);arr[adress] = new Node1074(adress, data, next);}int temp = adressStart;List<Node1074> list = new ArrayList<Node1074>();while(temp != -1) {//按链表顺序存入list中list.add(arr[temp]);temp = arr[temp].next;}int count = list.size() / before;//求需要反转几次List<Node1074> list2 = new ArrayList<Node1074>();for(int j = 1; j <= count ;j++) {//反转次数for(int i = before * j - 1;i >= before*(j - 1) ;i--) {list2.add(list.get(i));}}for(int i = before * count; i < list.size();i++) {//将剩余元素加入list2中list2.add(list.get(i));}temp = -1;for(int i = list2.size() - 1;i >= 0;i--) {//调整地址list2.get(i).next = temp;temp = list2.get(i).adress;;}for(Node1074 data : list2) {//输出System.out.println(data);}}
}
class Node1074{int adress;int data;int next;public Node1074(int adress, int data, int next) {this.adress = adress;this.data = data;this.next = next;}@Overridepublic String toString() {if(this.next != -1)return String.format("%05d", adress) + " " + data + " " + String.format("%05d", next);elsereturn String.format("%05d", adress) + " " + data + " " + String.format("%02d", next);}
}

PAT提交截图：