pat 1074 Reversing Linked List (25分) 链表反转

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:

?

1 2 3 4 5 6 7

00100 6 4 00000 4 99999 00100 1 12309 68237 6 -1 33218 3 00000 99999 5 68237 12309 2 33218

Sample Output:

?

1 2 3 4 5 6

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

?

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45

#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\n",V[i].data,V[i].order,V[i+1].data);     }           printf("%05d %d %d\n",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