PAT A1074 Reversing Linked List （25 分）——链表，vector，stl里的reverse

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 (≤10​5​​) 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

 

#include <stdio.h>
#include <map>
#include <algorithm>
#include <iostream>
#include <string>
#include <math.h>
#include <vector>
using namespace std;
const int maxn = 100010;
struct node{
    int addr;
    int data;
    int next;
}nodes[maxn];
vector<node> v;
int main(){
    int st, n, k, count = 0;
    cin >> st >> n >> k;
    for (int i = 0; i < n; i++){
        int start, data, next;
        cin >> start >> data >> next;
        nodes[start].addr = start;
        nodes[start].data = data;
        nodes[start].next = next;
    }
    while (st != -1){
        v.push_back(nodes[st]);
        st = nodes[st].next;
        count++;
    }
    for (int i = 0; i+k <= count; i = i + k){
        reverse(v.begin() + i, v.begin() + i + k);
    }
    for (int i = 0; i < count-1; i++){
        v[i].next = v[i + 1].addr;
    }
    v[count-1].next = -1;
    for (int i = 0;i < count-1; i++){
        printf("%05d %d %05d
", v[i].addr, v[i].data, v[i].next);
        
    }
    printf("%05d %d %d
", v[count - 1].addr, v[count - 1].data, v[count - 1].next);
    system("pause");
}

注意点：同B1025，链表题，都会要你重新排序输出，注意可以用vector实现链表，不要用静态数组，保存节点时一定要保存自己的地址。反转可以直接使用algorithm里的reverse