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 (≤) which is the total number of nodes, and a positive K (≤) 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
解题思路：
这道题目，用一个vector，按照顺序把 节点依次存起来，输入是4 - > 1 - > 6 -> 3 ->5 -> 2 按照 1->2->3->4->5->6存放，再反转vector
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<vector>
#include<algorithm>
using namespace std;
typedef struct Node{
int data;
int order;
int next;
}Node;
Node nodes[100010];
int main(){
// freopen("C:\Users\zzloyxt\Desktop\1.txt","r",stdin);
int first,n,k;
scanf("%d %d %d",&first,&n,&k);
for(int i=0;i<n;i++){
int data,order,next;
scanf("%d %d %d",&data,&order,&next);
nodes[data].data = data;
nodes[data].order = order;
nodes[data].next = next;
}
vector<Node> V;
for(int i=first;i!=-1;i = nodes[i].next){
V.push_back(nodes[i]);
}
for(int i=0;i<=V.size()-k;i=i+k){
reverse(V.begin()+i,V.begin()+i+k);
}
for(int i=0;i<V.size()-1;i++){
printf("%05d %d %05d
",V[i].data,V[i].order,V[i+1].data);
}
printf("%05d %d %d
",V[V.size()-1].data, V[V.size()-1].order,-1);
return 0;
}
注意这里的reverse()函数，传的是iterator，reverse(V.begin(), V.begin() + k)，最后一个单独输出