【PAT A1074】Reversing Linked List

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.

思路与样例
我的思路比较笨，就是模拟链表，正好熟悉下链表操作，注释应该很详细了。这里主要给出一个样例，你先可以试试书上的样例，如果不行再看我的，因为我书上的样例能过但是有一个答案错误。看到网上比较聪明的方法就是，按照链表顺序把节点放到一个vector或者数组里面，然后用头文件algorithm里的reverse函数一节一节地翻转就完事儿了，我就不贴代码了。

11111 5 4
11111 1 22222
22222 3 33333
33333 4 -1
44444 5 55555
55555 2 11111

代码

#include <cstdio>
using namespace std;

struct Node{
	int data, Next;
}Nodes[100010];

int main(){
	int start, N, K;
	scanf("%d%d%d", &start, &N, &K);
	for(int i = 0; i < N; i++){
		int address;
		scanf("%d", &address);
		scanf("%d", &Nodes[address].data);
		scanf("%d", &Nodes[address].Next);
	}

	//cnt记录已经完成操作的节点个数，K个为一组，以为倍数
	//begin为组的开始下标
	//last是上一组的第一个元素
	int cnt = 0, begin = start, end = start, last;
	//还有没有完成操作的组
	while(true){
		//iterator用来遍历链表，before用来记录原序列中当前节点的前一个节点
		int iterator = begin, before = begin;
		//找这一组的最后一个节点
		end = begin;
		for(int i = 0; i < K - 1 && end != -1; i++)
			end = Nodes[end].Next;
		//如果end是-1说明这一组个数不足K，此时不调整顺序
		if(end == -1)
			break;
		//上一组最后一个节点连接这一组调整后的第一个节点
		//从第二组开始
		if(cnt > 0)
			Nodes[last].Next = end;
		last = begin;//记录这一组调整后的最后一个元素
		//第一组的最后一个元素为整个链的首节点
		if(cnt == 0)
			start = end;
		//修改begin为下一组第一个节点
		begin = Nodes[end].Next;
		for(int i = 0; i < K; i++){
			int tmp = Nodes[iterator].Next;//用来保存修改next之前的next
			if(i != 0)
				Nodes[iterator].Next = before;
			before = iterator;
			iterator = tmp;
		}
		cnt += K;
	}
	//防止最后一组不用调整，此时上一组的最后节点接该组第一个节点（未调整）
	if(cnt > 0)
		Nodes[last].Next = begin;

	for(int i = 0; i < N && start != -1; i++){
		if(i < N - 1 && Nodes[start].Next != -1)
			printf("%05d %d %05d\n", start, Nodes[start].data, Nodes[start].Next);
		else
			printf("%05d %d -1\n", start, Nodes[start].data);
		start = Nodes[start].Next;
	}

	return 0;
}