1074 Reversing Linked List（链表逆序）

1074 Reversing Linked List （25 分）

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

分析

本题考查链表的逆序，注意边界条件。

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
const int N=100010;
int ad_val[N],val_ad[N],nextAddr[N];
vector<int>v;
int main() {
    int addr,n,k;
    cin>>addr>>n>>k;
    for(int i=0; i<n; i++) {
        int ad,val,nxt;
        cin>>ad>>val>>nxt;
        ad_val[ad]=val;
        val_ad[val]=ad;
        nextAddr[ad]=nxt;
    }
    for(int i=addr; i!=-1; i=nextAddr[i])
        v.push_back(ad_val[i]);
    for(int i=0; i+k<=(int)v.size(); i=i+k)
        reverse(v.begin()+i,v.begin()+i+k);
    for(int i=0; i<(int)v.size()-1; i++)
        printf("%05d %d %05d\n",val_ad[v[i]],v[i],val_ad[v[i+1]]);
    printf("%05d %d -1\n",val_ad[v[v.size()-1]],v[v.size()-1]);
        return 0;
}