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+10;
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=0; i<n; i++)
    {
        scanf("%d", &address);
        link[address].addr=address;
        scanf("%d%d", &link[address].data,&link[address].next);
    }
    int pos=0;
    stack<int> re;
    vector<int> ans;
    while(first != -1)
    {
        re.push(first);
        if(++pos == k)
        {
            pos=0;
            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=0; i<ans.size(); i++)
    {
        if(i!=0)
            printf("%05d\n", ans[i]);
        printf("%05d %d ", ans[i],link[ans[i]].data);
    }
    printf("-1");

    return 0;
}