1074. Reversing Linked List (25)

题目例如以下：

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

---

这是一道著名坑题，假设真的用单向链表来做，实在是太变态了，以下给出一种利用map和vector完毕的简单方法。

这道题最大的坑在于有无效结点，也就是说除了从头结点走到-1的所有结点外。还有其它子链表，排除的方法非常easy，仅仅要从头到尾走一遍记录下来，其它的所有扔掉。

要解决问题，须要考虑下面几个方面：

①怎样依据题目输入存储结点？

採用结构体数组。规模为10万，下标即为自己的地址。结构体内存储编号和下一个地址。

②怎样进行反转？

仅仅用vector记录有效结点的编号。同一时候在记录时用map记录编号到地址的相应关系。

仅仅反转vector中的编号。

③怎样连接地址输出？

通过反转后的vector。结合map查询地址。实现地址连接。

一定要注意处理K=N和N=1的情况，注意地址的前导0，-1不能有前导0。

代码例如以下：

#include <iostream>
#include <stdio.h>
#include <map>
#include <vector>

using namespace std;

struct Node{
    int me;
    int num;
    int next;
}nodes[100000];

int main()
{
    int N;
    map<int,int> addMap;
    int add,num,next;
    int head,K;
    cin >> head >> N >> K;
    for(int i = 0; i < N; i++){
        scanf("%d%d%d",&add,&num,&next);
        nodes[add].me = add;
        nodes[add].num = num;
        nodes[add].next = next;
        addMap[num] = add;
    }
    vector<int> validList;
    add = head;
    while(add != -1){
        Node n = nodes[add];
        validList.push_back(n.num);
        add = n.next;
    }
    vector<int> reverseList;
    int len = validList.size();
    int cur = len;
    for(int group = 0; group * K < len; group++){
        cur = group * K;
        if(len - cur < K) break;
        else{
            cur = len;
            for(int i = K - 1; i >= 0; i--){
                reverseList.push_back(validList[group * K + i]);
            }
        }

    }
    for(int i = cur; i < len; i++){
        reverseList.push_back(validList[i]);
    }
    len = reverseList.size();
    if(len == 1){
        int num = reverseList[0];
        printf("%05d %d -1
",addMap[num],num);
    }else{
        for(int i = 0; i < len - 1; i++){
            int now = reverseList[i];
            int next = reverseList[i+1];
            printf("%05d %d %05d
",addMap[now],now,addMap[next]);
        }
        int now = reverseList[len - 1];
        printf("%05d %d -1
",addMap[now],now);
    }
    return 0;
}