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 (<= 10^5^) 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
题目大意：给出n个节点，以及头结点，要求把相邻的k个节点翻转， 如果n不能被k整除，剩下的节点按照原来的顺序
思路：先用vector<int> value保存当前节点的值， vector<int> v保存当前节点指向下一个节点的指针； 然后再根据链表信息把k个链表的信息保存在vector<node> temp中， 然后在反向添加到vector<node> link中；
修改link中链表信息，输出结果
注意点：在使用vector的时候，如果事先申明了vector的大小，如果再使用push_back()在压在申请内存的后面的，而不是从头开始压入的；
此外，pat的链表类题中，经常存在不在链表中的节点， 所以在最后的循环输出的时候，不能以节点个数作为循环结束的判断，应该以link的大小作为循环次数的判断
1 #include<iostream> 2 #include<vector> 3 using namespace std; 4 struct node{ 5 int addr, val, next; 6 }; 7 8 int main(){ 9 int root, n, k, i, j; 10 cin>>root>>n>>k; 11 vector<int> value(100000), v(100000); 12 int addr, val, next; 13 for(i=0; i<n; i++){ 14 scanf("%d %d %d", &addr, &val, &next); 15 value[addr] = val; 16 v[addr] = next; 17 } 18 vector<node> link; 19 for(i=0; i<n/k+1; i++){ 20 node nnode; 21 vector<node> temp; 22 for(j=0; j<k; j++){ 23 if(root==-1) break; 24 nnode.addr=root; 25 nnode.val=value[root]; 26 nnode.next=v[root]; 27 root=v[root]; 28 temp.push_back(nnode); 29 } 30 if(temp.size()==k) for(j=k-1; j>=0; j--) link.push_back(temp[j]); 31 else for(j=0; j<temp.size(); j++) link.push_back(temp[j]); 32 } 33 34 for(i=0; i<link.size(); i++){ 35 if(i!=link.size()-1){ 36 link[i].next=link[i+1].addr; 37 printf("%05d %d %05d\n", link[i].addr, link[i].val, link[i].next); 38 }else printf("%05d %d %d", link[link.size()-1].addr, link[link.size()-1].val, -1); 39 } 40 41 return 0; 42 }