1074. Reversing Linked List (25)

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.

00100 6 400000 4 9999900100 1 1230968237 6 -133218 3 0000099999 5 6823712309 2 33218Sample Output:
00000 4 3321833218 3 1230912309 2 0010000100 1 9999999999 5 6823768237 6 -1
#include<stdio.h>#include<algorithm>#include<stack>using namespace std;struct Node{	int data;	int next;}node[100000];//定義靜態(tài)鏈表節(jié)點(diǎn)struct record{	int address;	//int next;};//用于反轉(zhuǎn)鏈表int main(){	for (int i = 0; i < 100000; i++)	{		node[i].data = 0;		node[i].next = -1;	}//初始化	int head, N, K;	scanf("%d%d%d", &head, &N, &K);	int address, data, next;	while (N--)	{		scanf("%d%d%d", &address, &data, &next);		node[address].data = data;		node[address].next = next;//鏈接各節(jié)點(diǎn)	}	int count = 0;//計(jì)數(shù)器	int addressNow = head;	int pre;	stack<record> sta;//用于暫存要被逆轉(zhuǎn)連接的節(jié)點(diǎn)	record temprecord;//保存節(jié)點(diǎn)地址,本來以為要保存下一個(gè)節(jié)點(diǎn)的地址，其實(shí)根本不需要，也不想改了就這樣吧	int oldLast;//上一次逆轉(zhuǎn)后指向最后一個(gè)節(jié)點(diǎn)的指針	while (addressNow!= -1)	{		temprecord.address = addressNow;		//temprecord.next = node[addressNow].next;		sta.push(temprecord);		pre = addressNow;		addressNow = node[addressNow].next;		count++;		if (count%K == 0)		{					if (count == K)					head = pre;//第一個(gè)開始反轉(zhuǎn)的節(jié)點(diǎn)成為頭了				if (!sta.empty())				{					sta.pop();				}				if (pre != head)				{					node[oldLast].next = pre;				}//如果不是第一個(gè)那么需要將上一次反轉(zhuǎn)后的最后一個(gè)節(jié)點(diǎn)的下個(gè)地址做修改				//將樣例中的K由4改為2你就知道為什么了			for (int i = 0; i < K-1; i++)			{				node[pre].next = sta.top().address;//逆轉(zhuǎn)鏈表				pre = sta.top().address;				if (!sta.empty())				{					sta.pop();				}			}			node[pre].next = addressNow;//修改反轉(zhuǎn)節(jié)點(diǎn)的最后一個(gè)節(jié)點(diǎn)的下一個(gè)地址，			//若后面不會(huì)再反轉(zhuǎn)，那么就是當(dāng)前掃描的節(jié)點(diǎn)的地址，注意我是先計(jì)數(shù)后又把指針			//往后移一位了			oldLast = pre;//保存反轉(zhuǎn)節(jié)點(diǎn)的最后一個(gè)節(jié)點(diǎn)的地址		}	}	addressNow = head;	while (addressNow!= -1)	{		if(node[addressNow].next!=-1)		printf("%05d %d %05d/n", addressNow, node[addressNow].data, node[addressNow].next);		else			printf("%05d %d %d/n", addressNow, node[addressNow].data, node[addressNow].next);		addressNow = node[addressNow].next;	}	return 0;}/*注意把每次反轉(zhuǎn)后的節(jié)點(diǎn)的下一個(gè)地址修改正確，比如說1 2 3 4 5 6，反轉(zhuǎn)個(gè)數(shù)K為2時(shí)，應(yīng)該是2 1 4 3 6 5，這時(shí)要記得把1跟4連上，3跟6連上。*/