topic
Given a constant KK K and a singly linked list LL L , please write a program to reverse every KK K nodes in LL L. For example: Given that LL L is 1→2→3→4→5→6 and KK K is 3, the output should be 3→2→1→6→5→4; if KK K is 4, the output should be 4→3→2→1→5→6, that is, less than KK K elements are not inverted at the end.
Input format:
Each input contains 1 test case. Each test case is given first line of the address of a node, the total number of positive integers node NN N ( ≤ 1 0. 5 \ ^ Le. 5 10 ≤ 1 0 ? . 5 ? ? ), And a positive integer KK K ( ≤ N \le N ≤ N ), that is, the number of sub-links that require reversal. The address of the node is a 5-bit non-negative integer, and the NULL address is represented by ? 1 -1 ? 1 .
Next, there are NN N lines, each in the format:
Address Data Next
Wherein Address a node address, Data is stored in the node integer data, Next the address of the next node.
Output format:
For each test case, the reversed linked list is output in order, each node on it occupies a row, and the format is the same as the input.
Input sample:
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
Code
#include<iostream>
#include<algorithm>
#include<string>
using namespace std;
string format(int i)
{
if (i < 10)
return "0000" + to_string(i);
else if (i < 100)
return "000" + to_string(i);
else if (i < 1000)
return "00" + to_string(i);
else if (i < 10000)
return "0" + to_string(i);
else if (i < 100000)
return to_string(i);
}
int main()
{
int BEGIN, n, K, i,addr,content,next,sum=0;
cin >> BEGIN >> n >> K;
int S[100005][2];
for (i = 0; i < 100005; i++)
{
S[i][0] = 0;
S[i][1] = 0;
}
i = -1;
while (++i < n)
{
cin >> addr >> content >> next;
S[addr][0] = content;
S[addr][1] = next;
}
int** a = new int* [n];
for (i = 0; i < n; i++)
a[i] = new int[2];
while (BEGIN != -1)
{
a[sum++][0] = BEGIN;
a[sum-1][1] = S[BEGIN][0];
BEGIN = S[BEGIN][1];
}
n = sum;
if (K > 1)
{
for (i = 0; i < n; i += K)
if (i + K <= n)
reverse(a + i, a + i + K);
}
for (i = 0; i < n; i++)
if (i + 1 != n)
cout << format(a[i][0]) << " " << a[i][1] << " " << format(a[i + 1][0]) << endl;
else
cout << format(a[i][0]) << " " << a[i][1] << " " << "-1" << endl;
return 0;
}