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 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
这道题之前做过，乙级里有，当时不会做，参考写的，学到了，这次一看到这个题就有了思路，只希望不是定式思维就好，第一次提交就过了两个测试点，4分。。第二次才满分，原因在于最后的输出那里，应该是v[i+1]，我给弄成了v[i]的next，就错了，有的结点不一定在链表上，这是这个题的一个坑，之前做就觉得很坑，唉这次还是被坑了，所以需要记录有效的个数，这里由于采用的vector，直接用它的大小记录就可以，不用计数器。下边这个链接题的做法和这个的思路一样，可以对比着写
#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
typedef struct{
int data;
int next;
}Node;
Node node[100005];
vector<int> v;
int main(){
ios::sync_with_stdio(false);
int head,n,k;
cin >> head >> n >> k;
for(int i = 0;i<n;i++){
int add;
cin >> add;
cin >> node[add].data >> node[add].next;
}
while(head!=-1){
v.push_back(head);
head = node[head].next;
}
int len = v.size();
for(int i = 0;i<(len-len%k);i+=k){
reverse(v.begin()+i,v.begin()+i+k);
}
for(int i = 0;i<len-1;i++){
printf("%05d %d %05d\n",v[i],node[v[i]].data,v[i+1]);
}
printf("%05d %d -1\n",v[len-1],node[v[len-1]].data);
return 0;
}