1074. 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

用数组存链表，地址即为下标，按照地址读一遍链表，然后存在指针数组p中，再按要求输出。题目肯定不是考验这个所以还是应该把链表按要求反转了再输出。
代码：

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <map>
using namespace std;
struct node
{
    int add,data,next;
}s[100000],*p[100000],*temp = NULL;
int main()
{
    int n,k,address,a,b,c,d = 0;
    scanf("%d%d%d",&address,&n,&k);
    for(int i = 0;i < n;i ++)
    {
        scanf("%d %d %d",&a,&b,&c);
        s[a].add = a;
        s[a].data = b;
        s[a].next = c;
    }
    for(int i = address;i != -1;i = s[i].next)
    {
        p[d ++] = &s[i];
    }
    for(int i = 0;i < d;i += k)
    {
        if(d - i >= k)
        {
            if(temp != NULL)printf("%05d %d %05d\n",temp -> add,temp -> data,p[i + k - 1] -> add);
            for(int j = k - 1;j > 0;j --)
            {
                printf("%05d %d %05d\n",p[i + j] -> add,p[i + j] -> data,p[i + j - 1] -> add);
            }
            temp = p[i];
            if(d - i == k)printf("%05d %d %d\n",p[i] -> add,p[i] -> data,-1);
        }
        else
        {
            if(temp != NULL)printf("%05d %d %05d\n",temp -> add,temp -> data,p[i] -> add);
            for(int j = i;j < d;j ++)
            {
                if(j == d - 1)printf("%05d %d %d\n",p[j] -> add,p[j] -> data,p[j] -> next);
                else printf("%05d %d %0d\n",p[j] -> add,p[j] -> data,p[j] -> next);
            }
        }
    }
}

将链表按要求反转。

代码：

#include <stdio.h>
struct node {
    int data,next;
}s[100000],*t,*p,*q,*r,*head,*tail = NULL;
int main() {
    int n,k,address,a,b,c;
    scanf("%d%d%d",&address,&n,&k);
    if(k > n) k = n;
    for(int i = 0;i < n;i ++) {
        scanf("%d %d %d",&a,&b,&c);
        s[a].data = b;
        s[a].next = c;
    }
    int nn = 0;
    for(int i = address;i != -1;i = s[i].next) nn ++;
    q = &s[address];
    while(k <= nn) {
        t = p = q;
        q = &s[q -> next];
        c = 1;
        while(c < k) {
            r = q;
            if(q -> next != -1) q = &s[q -> next];
            r -> next = t - s;
            t = r;
            c ++;
        }
        if(tail == NULL) head = t,tail = p;
        else tail -> next = t - s,tail = p;
        nn -= k;
    }
    if(nn % k == 0) tail -> next = -1;
    else tail -> next = q - s;
    address = head - s;
    while(address != -1) {
        if(s[address].next == -1) printf("%05d %d %d\n",address,s[address].data,s[address].next);
        else printf("%05d %d %05d\n",address,s[address].data,s[address].next);
        address = s[address].next;
    }
    return 0;
}

来源：oschina

链接：https://my.oschina.net/u/4261514/blog/4197667

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