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 33218Sample Output:
00000 4 33218 33218 3 12309 12309 2 00100 00100 1 99999 99999 5 68237 68237 6 -1
用c语言，用数组存链表，地址即为下标，按照地址读一遍链表，然后存在指针数组p中，再按要求输出。
代码：
#include <cstdio> #include <iostream> #include <algorithm> #include <cstring> #include <map> using namespace std; struct node { int add,data,next; }s[100000],*p[100000],*temp = NULL; int main() { int n,k,address,a,b,c,d = 0; scanf("%d%d%d",&address,&n,&k); for(int i = 0;i < n;i ++) { scanf("%d %d %d",&a,&b,&c); s[a].add = a; s[a].data = b; s[a].next = c; } for(int i = address;i != -1;i = s[i].next) { p[d ++] = &s[i]; } for(int i = 0;i < d;i += k) { if(d - i >= k) { if(temp != NULL)printf("%05d %d %05d\n",temp -> add,temp -> data,p[i + k - 1] -> add); for(int j = k - 1;j > 0;j --) { printf("%05d %d %05d\n",p[i + j] -> add,p[i + j] -> data,p[i + j - 1] -> add); } temp = p[i]; if(d - i == k)printf("%05d %d %d\n",p[i] -> add,p[i] -> data,-1); } else { if(temp != NULL)printf("%05d %d %05d\n",temp -> add,temp -> data,p[i] -> add); for(int j = i;j < d;j ++) { if(j == d - 1)printf("%05d %d %d\n",p[j] -> add,p[j] -> data,p[j] -> next); else printf("%05d %d %0d\n",p[j] -> add,p[j] -> data,p[j] -> next); } } } }
