数据结构练习 02-线性结构2. Reversing Linked List (25)

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

提交了10次才完全正确，花了快5个小时。感觉自己实力好差。

#include<iostream>
#include <string>
using namespace std;
class linked{
public:
    int data;
    int next;
};
linked *list = new linked[100000];
int prehead;
int nexthead;
int Reverse(int head, int k){
    int cnt = 1,temp;
    int news = head;
    int olds = list[head].next;
    while (cnt < k){
        temp = list[olds].next;
        list[olds].next = news;
        news = olds;
        olds = temp;
        cnt++;
    }
    list[head].next = olds;
    prehead = head;
    nexthead = olds;
    return news;
}
int main(){
    int firstNode, address, next,N, k, data,head,preshead;
    int sum = 0;
    cin >> firstNode >> N >> k;
    for (int i = 0; i < N; i++){
        cin >> address >> data >> next;
        list[address].data = data;
        list[address].next = next;
    }
    int p = firstNode;
    while (p != -1){
        sum++;
        p = list[p].next;
    }
    head = Reverse(firstNode, k);
    for (int i = 1; i < sum / k; i++){
        preshead = prehead;
        list[preshead].next = Reverse(nexthead, k);
    }
    p = head;
    while (p!= -1){
        printf("%05d ",p);
        cout << list[p].data <<" ";
        if (list[p].next != -1)
            printf("%05d\n", list[p].next);
        else
            cout << list[p].next << endl;
        p = list[p].next;
    }

    return 0;
}

ac的图片看着真的很爽。