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 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
#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
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。
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