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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 #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 }
02-1 Reversing Linked List （PAT）
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原文地址：http://www.cnblogs.com/liangchao/p/4271129.html
