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?Lbeing 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?(≤10?5??) 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

總結：

學習了柳神的代碼，for循環里的真漂亮

代碼：

#include <iostream> 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; }

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總結

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