1165 Block Reversing (25 分) PAT

Given a singly linked list L. Let us consider every K nodes as a block (if there are less than K nodes at the end of the list, the rest of the nodes are still considered as a block). Your job is to reverse all the blocks in L. For example, given L as 1→2→3→4→5→6→7→8 and K as 3, your output must be 7→8→4→5→6→1→2→3.

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 size of a block. 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 8 3
71120 7 88666
00000 4 99999
00100 1 12309
68237 6 71120
33218 3 00000
99999 5 68237
88666 8 -1
12309 2 33218
 

Sample Output:

71120 7 88666
88666 8 00000
00000 4 99999
99999 5 68237
68237 6 00100
00100 1 12309
12309 2 33218
33218 3 -1
 

这题是很基础的链表问题，对于这种问题，我一般喜欢写链式前向星，不喜欢指针形式，复杂且指的头疼。

直接先全部翻转，然后分组翻转即可。

注意：最后一部分不足k的话也是需要翻转的。

#include 
#include 
#include 
#include 

using namespace std;

const int N = 100010;

int n, k;
int h, e[N], ne[N];

int main()
{
    scanf("%d%d%d", &h, &n, &k);

    for (int i = 0; i < n; i ++ )
    {
        int address, data, next;
        scanf("%d%d%d", &address, &data, &next);
        e[address] = data, ne[address] = next;
    }

    vector q;
    for (int i = h; i != -1; i = ne[i]) q.push_back(i);
    int size = q.size() % k; // 这一步必须思考清晰,取余则是多余数据
    // size / k 则是你有几个block
    reverse(q.begin(), q.end());
    bool flag = true;
    for(int i = 0; i < q.size(); )
    {
        if (flag) reverse(q.begin() + i, q.begin() + i + size), i += size, flag = false;
        else 
        {
            reverse(q.begin() + i, q.begin() + i + k);
            i += k;
        }
    }
    for (int i = 0; i < q.size(); i ++ )
    {
        printf("%05d %d ", q[i], e[q[i]]);
        if (i + 1 == q.size()) puts("-1");
        else printf("%05dn", q[i + 1]);
    }

    return 0;
}