1025. The inversion list (25)
Given a constant K and a singly linked list L, please write a program to L in each reverse K nodes, such as: given the L for 1-2-3-4-5-6, K is 3, the output should be 3-2-1-6-5 -> 4; If is 4 K, the output should be 4-3-2-1-5-6, last less than K element does not reverse,

Input format:

Each input contains one test case, each test case line 1 give the address of the first node, the total number of positive integer N (& lt;=105), as well as the positive integer K (& lt;=N), which requires inversion of sub chain node number, the address of the node is five nonnegative integers, NULL address expressed with 1,

The next N lines, each line format for:

The Address Data Next

The Address is a node Address, the Data is saved the nodes integer Data, Next is the Address of the Next node,

The output format:

Sequence of each test case, output the list after inversion, each node of a line, its format is the same as the input,

Input the sample:
00100 June 4
00000 4 99999
00100 1 12309
68237 6-1
33218 3 00000
99999 5 68237
12309 2 33218
The output sample:
00000 4 33218
33218 3 12309
12309 2 00100
00100 1 99999
99999 5 68237
68237 6-1

Here is my code:

 
# include 
using namespace std; 
Struct Node {
Int AD; 
The int data; 
Int next; 
} node [100001], \ [100001]; 
Int main () 
{
Int fstad, n, k; 
While (the scanf (" % d % d % d ", & amp; Fstad, & amp; N, & amp; K)!=(EOF) {//input starting address, number of nodes and exchange length 
for(int i=0; IThe scanf (" % d % d % d ", & amp; The node [I]. AD, & amp; The node [I]. The data, & amp; The node [I] next);//the address of the input node, the node number and point to the next node address 
} 
Int j=0;//statistical effective number of node 
for(int i=0; IIf (fstad==node [I] AD) {
\ [j++]=node [I]; 
Fstad=node [I] next; 
I=1; 
} 
If (fstad==1) break; 
} 
Int cont=0; 
Int t=0; 
For (int I=k - 1; I>=0 & amp; & ContNode [t++]=\ [I]; 
If (I==cont * k) {
Cont++; 
I=cont * k + k; 
} 
} 
If (j % k! Enough K=0) {//behind the part directly into the corresponding node in the [] 
For (int I=j - j % k; INode [I]=\ [I]; 
} 
} 
for(int i=0; IThe node [I] next=node [I + 1] AD; 
} 
for(int i=0; IIf (node [I] next==1) printf (" % 5 d % d % d \ n ", the node [I] AD, node [I] data, node [I] next); 
The else printf (" % 5 d % d % 5 d \ n ", the node [I] AD, node [I] data, node [I] next); 
} 
} 
return 0; 
} 

I tested several examples, can not have a problem, but OJ judgment is the answer error and timeout, help everybody to help, thank you,

CodePudding user response:

@ confident boy can help have a look, thank you

CodePudding user response:

"If it is 4 K output should be 4-3-2-1-5-6"
If this sentence, K is 4, the output should be: 4-3-2-1-6 to 5

CodePudding user response:

@ confident boy subject to output the 4-3-2-1-5-6, also is the final can not meet the needs of K to transpose

CodePudding user response:

I also am the mistake, the building Lord found the problem?