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
和乙级的1025一样:1025 反转链表 (25 分)_Brosto_Cloud的博客-CSDN博客
Data 和 Next 数组的下标 i 都是地址,表示对应地址的数据和链接的下一个地址,list数组下标 i 是顺序,代表节点地址,每个地址对应的数据都是不变的,改变的只是链接的下一个地址,所以对list 数组每k个进行反转后直接输出 list[i]和list[i+1]就是对应的地址顺序,data[i]是对应的数据。
#include <iostream>
#include <algorithm>
#include <iomanip>
using namespace std;
int Data[100010], list[100010], Next[100010];
int n, k, addr, t, sum;
int main() {
cin >> addr >> n >> k;
for (int i = 0; i < n; i++) {
cin >> t;
cin >> Data[t] >> Next[t];
}
while (addr != -1) {
list[sum++] = addr;
addr = Next[addr];
}
for (int i = 0; i < (sum - sum % k); i += k) {
reverse(list + i, list + i + k);
}
for (int i = 0; i < sum - 1; i++) {
cout << setfill('0') << setw(5) << list[i] << ' ' << Data[list[i]] << ' ' << setfill('0') << setw(
5) << list[i + 1] << endl;
}
cout << setfill('0') << setw(5) << list[sum - 1] << ' ' << Data[list[sum - 1]] << ' ' << -1;
return 0;
}