02-线性结构3 Reversing Linked List (25分)

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给定一个正整数K和一个单链表L，你需要反转链表中每隔K个元素的链接。例如，对于链表1→2→3→4→5→6，如果K=3，则输出3→2→1→6→5→4；如果K=4，则输出4→3→2→1→5→6。输入包含链表的首节点地址、节点总数N和要反转的子列表长度K。输出反转后的链表，每个节点按输入格式打印。

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 $N$ ( $\le 10^5$ ) which is the total number of nodes, and a positive $K$ ( $\le N$ ) 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 Nextis 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：主要利用C++的reverse函数，还有C++的cout要格式化输出没有printf方便

C++代码：

#include <iostream>
#include <algorithm>
#include <vector>
#include <stdio.h>
using namespace std;
#define maxlen 100001

struct node{
	int add;
	int data;
	int next;
};
typedef struct node Node;
Node nodes[maxlen];
vector<Node> nodeList;
int main(){
	int firstAdd, N, K;
	cin>>firstAdd>>N>>K;
	while(N--){
		Node n;
		cin>>n.add>>n.data>>n.next;
		nodes[n.add] = n;
	}
	int address = firstAdd;
	while(address != -1){
		nodeList.push_back(nodes[address]);
		address = nodes[address].next;
	}
	//到此nodeList存储的就是顺序的列表
	int len = nodeList.size();
	int period = len/K;
	for(int i = 1; i <= period; i++){
		int head = (i-1)*K;
		int end = i*K;
		reverse(nodeList.begin()+head, nodeList.begin()+end);
	}
	//reverse complete
	//print nodeList
	for(int i = 0; i < len-1; i++){
		//cout<<nodeList[i].add<<" "<<nodeList[i].data<<" "<<nodeList[i+1].add<<endl;
		printf("%05d %d %05d\n", nodeList[i].add, nodeList[i].data, nodeList[i+1].add);
	}
	//cout<<nodeList[len-1].add<<" "<<nodeList[len-1].data<<" "<<-1;
	printf("%05d %d %d\n",nodeList[len-1].add, nodeList[len-1].data, -1);
	return 0;
}

2017-4-16 待续

00000 4 33218
33218 3 12309
12309 2 00100
00100 1 99999
99999 5 68237
68237 6 -1