1074. Reversing Linked List (25)-PAT甲级真题

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

分析：把地址为temp的数的数值存入data[temp]中，把temp的下一个结点的地址存入next[temp]中。

注意：不一定所有的输入的结点都是有用的，加个计数器sum

#include 
using namespace std;
int main() {
    int first, k, n, sum = 0;
    cin >> first >> n >> k;
    int temp, data[100005], next[100005], list[100005], result[100005];
    for (int i = 0; i < n; i++) {
        cin >> temp;
        cin >> data[temp] >> next[temp];
    }
    while (first != -1) {
        list[sum++] = first;
        first = next[first];
    }
    for (int i = 0; i < sum; i++) result[i] = list[i];
    for (int i = 0; i < (sum - sum % k); i++)
        result[i] = list[i / k * k + k - 1 - i % k];
    for (int i = 0; i < sum - 1; i++)
        printf("%05d %d %05d\n", result[i], data[result[i]], result[i + 1]);
    printf("%05d %d -1", result[sum - 1], data[result[sum - 1]]);
    return 0;
}