Reversing Linked List （java实现）

7-2 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 $N$ ( $\leq 1 0^{ 5 }$ ) which is the total number of nodes, and a positive $K$ ( $\leq N$ ) 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

思路：

1、用java中的LinkedList类建立链表，先按照顺序串起来。

2、翻转顺序搞清楚，先用temp节点copy第k个节点，然后从后到前依次翻转。

3、这题目有个坑，就最后一个测试点会给你额外的链表外的节点，要把这些节点剔除出去。

4、java跟c写同样的程序，c出结果只要2ms，java要100+ms，所以倒数第二个测试点，当数据量最大时，c需要100+ms，题目允许最大运行时间是400ms，不用说，java运行超时了。

java：

测试点	结果	耗时	内存
0	答案正确	137 ms	23520KB
1	答案正确	133 ms	22584KB
2	答案正确	136 ms	23696KB
3	答案正确	133 ms	23404KB
4	答案正确	129 ms	23296KB
5	运行超时	0 ms	0KB
6	答案正确	138 ms	23680KB


import java.util.Scanner;

// 翻转链表
public class Main {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner in = new Scanner(System.in);	
		Node head = new Node();
		head.naddr = in.nextInt();
		int n = in.nextInt();
		int k = in.nextInt();
		
		Node [] a = new Node[n+1];
		a[0]=head;
		
		for(int i = 1;i<=n;i++) {
			Node nd = new Node();
			nd.addr = in.nextInt();
			nd.value = in.nextInt();
			nd.naddr = in.nextInt();
			a[i] = nd;
		}
		
		int j = 0;
		for(int count = 0 ; count<n; count++) {
			for(int i = 0;i < n+1; i++ ) {
				if(a[i].addr == a[j].naddr)
					{
						a[j].next = a[i];
						j=i;
						break;
					}
			}
		}
		
		Node js = head;
		n = 0;
		while(js.naddr!=-1)
		{
			n++;
			js = js.next;
		}
		
//		print(head,n);
		
		Node result = reverse(head,n,k);
		
		print(result,n);
			
	}

	private static void print(Node result,int n) {
		// TODO Auto-generated method stub
		result = result.next;
		for(int i= 0 ; i<n;i++) {
			System.out.printf("%05d", result.addr);
			System.out.print(" "+result.value+" ");
			if(result.naddr!=-1) {
				System.out.printf("%05d", result.naddr);
				System.out.println();
			}
			else
				System.out.print(-1);
			
			result = result.next;
		}
	}

	private static Node reverse(Node head, int n, int k) {
		// TODO Auto-generated method stub
		
		int count = n/k;
		int rest = n%k;
		Node point = head;
		
		for( int i = 0; i<count ; i++) {
			Node temp = new Node();
			Node temp2 = get(point,k);
			temp.naddr = get(point,k).naddr;
			temp.next = get(point,k).next;
			
			for(int j = k ; j>=2 ; j--) {
				get(point,j).naddr = get(point,j-1).addr;
				get(point,j).next  = get(point,j-1);
				
//				System.out.println(get(point,j).naddr);
			}
			
			get(point,1).naddr = temp.naddr;
			get(point,1).next = temp.next;
			
			
			
			point.naddr = temp2.addr;
			point.next = temp2;
			
			point = get(point,k);

			
		}
		if(rest == 0 )
			point.naddr = -1;
		
		return head;
	}

	private static Node get(Node point, int k) {
		// TODO Auto-generated method stub
		for(int i =0; i<k; i++)
			point = point.next;
		return point;
	}

}
		class Node{
			public int addr = -1;
			public int value = -1;
			public int naddr = -1;
			public Node next = null;
		
			Node(){
				;
			}
	}