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
1.使用递归求解
2.考虑输入中包含孤立节点的情况
1 #include <iostream>
2 #include <string>
3 #include <stdlib.h>
4 #include <iomanip>
5 using namespace std;
6
7 typedef struct{
8 int data;
9 int next;
10 } dataNode;
11
12 int ReverseList( dataNode node[], int numNode, int lenSub, int firAdd );
13
14 int main()
15 {
16 //输入处理
17 int firAdd, numNode, lenSub;
18 //第一行处理
19 cin >> firAdd >> numNode >> lenSub;
20 dataNode nodes[100000];
21 int i;
22 int addr, data, next, head;
23 for ( i = 0; i < numNode; i++ )
24 {
25 cin >> addr >> data >> next;
26 nodes[ addr ].data = data;
27 nodes[ addr ].next = next;
28 }
29 //排除无效节点，得到有效节点数量
30 int noNode = firAdd;
31 int realNum = 0;
32 while( noNode != -1 )
33 {
34 noNode = nodes[ noNode ].next;
35 realNum++;
36 }
37 head = ReverseList( nodes, realNum, lenSub, firAdd );
38 while ( head != -1 )
39 {
40 cout << setw(5) << setfill(‘0‘) << head << ‘ ‘;
41 cout << nodes[ head ].data << ‘ ‘;
42 if ( nodes[ head ].next == -1 )
43 {
44 cout << nodes[ head ].next << endl;
45 }
46 else
47 {
48 cout << setw(5) << setfill(‘0‘) << nodes[ head ].next << endl;
49 }
50
51 head = nodes[ head ].next;
52 }
53 //输出处理
54 return 0;
55 }
56
57 int ReverseList( dataNode nodes[], int numNode, int lenSub, int firAdd )
58 {
59 if ( lenSub > numNode || nodes == NULL || firAdd == -1 ) return firAdd;
60 int current = firAdd;
61 int prev = -1;
62 int next = -1;
63 int count = 0;
64 while ( current != -1 && count < lenSub )
65 {
66 next = nodes[ current ].next;
67 nodes[ current ].next = prev;
68 prev = current;
69 current = next;
70 count++;
71 }
72 if ( next != -1 )
73 {
74 nodes[ firAdd ].next = ReverseList( nodes, numNode - lenSub, lenSub, next );
75 }
76 return prev;
77 }