PAT_A1074#Reversing Linked List

Source:

PAT A1074 Reversing Linked List (25 分)

Description:

Given a constant K and a singly linked list L, you are supposed to reverse the links of every Kelements 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

Keys:

• 栈和队列

Attention:

• vector<int> ans.insert(ans.begin()+pos,x);

Code:

 #include<cstdio>
 #include<stack>
 #include<vector>
 using namespace std;
 const int M=1e5+;
 struct node
 {
     int data;
     int addr,next;
 }link[M]; int main()
 {
 #ifdef ONLINE_JUDGE
 #else
     freopen("Test.txt", "r", stdin);
 #endif // ONLINE_JUDGE     int first,n,k,address;
     scanf("%d%d%d", &first,&n,&k);
     for(int i=; i<n; i++)
     {
         scanf("%d", &address);
         link[address].addr=address;
         scanf("%d%d", &link[address].data,&link[address].next);
     }
     int pos=;
     stack<int> re;
     vector<int> ans;
     while(first != -)
     {
         re.push(first);
         if(++pos == k)
         {
             pos=;
             while(!re.empty())
             {
                 ans.push_back(re.top());
                 re.pop();
             }
         }
         first = link[first].next;
     }
     int len=ans.size();
     while(!re.empty())
     {
         ans.insert(ans.begin()+len,re.top());
         re.pop();
     }
     for(int i=; i<ans.size(); i++)
     {
         if(i!=)
             printf("%05d\n", ans[i]);
         printf("%05d %d ", ans[i],link[ans[i]].data);
     }
     printf("-1");     return ;
 }