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.

Sample Input:

00100 6 400000 4 9999900100 1 1230968237 6 -133218 3 0000099999 5 6823712309 2 33218

Sample Output:

00000 4 3321833218 3 1230912309 2 0010000100 1 9999999999 5 6823768237 6 -1

#include<iostream>#include<map>using namespace std;struct Node{int address;int value;int next;Node(){ address = -1; value = -1; next = -1; }}*node;int main(){int head, num, k,current,tail,pre;map<int, int>addrtoIndex;scanf("%d%d%d",&head,&num,&k);node = new Node[num+1];int cnt = (num - num%k) / k;//add headNodenode[0].next = head;node[0].address = -2;addrtoIndex.insert(make_pair(-2, 0));for (int i = 1; i <= num; ++i){scanf("%d%d%d",&node[i].address,&node[i].value,&node[i].next);addrtoIndex.insert(make_pair(node[i].address,i));}current = head;pre = -2;for (int i = 0; i < cnt; ++i){int temp = current;for (int j = 0; j < k-1; ++j){tail = node[addrtoIndex[temp]].next;temp = tail;}for (int m = 0; m < k-1; ++m){node[addrtoIndex[pre]].next = node[addrtoIndex[current]].next;temp = node[addrtoIndex[tail]].next;node[addrtoIndex[tail]].next = current;node[addrtoIndex[current]].next = temp;current = node[addrtoIndex[pre]].next;}pre = current;current = node[addrtoIndex[current]].next;}current = node[addrtoIndex[-2]].next;for (int i = 0; i < num-1; ++i){printf("%05d %d %05d\n",node[addrtoIndex[current]].address, node[addrtoIndex[current]].value, node[addrtoIndex[current]].next);current = node[addrtoIndex[current]].next;}printf("%05d %d %d\n", node[addrtoIndex[current]].address, node[addrtoIndex[current]].value, node[addrtoIndex[current]].next);}