1074. Reversing Linked List (25)发布时间：2022/5/31 13:06:38
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 (5) which is the total number of nodes, and a positive K (
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
#include
#include
using namespace std;
const int maxn = 100100;
struct Node{
int address,data,next;
int order;
}node[maxn];
bool cmp(Node a,Node b){
return a.order b.order;
}
int main(){
int i,address;
for(i = 0; i ){
node.order = maxn;
}
int begin,n,k; //起始节点地址，节点数目，分组数
scanf("%d%d%d",&begin,&n,&k);
for(i = 0; i ){
scanf("%d",&address);
scanf("%d%d",&node[address].data,&node[address].next);
node[address].address = address;
}
int p = begin,count = 0;
while(p != -1){
node[p].order = count++;
p = node[p].next;
}
sort(node,node+maxn,cmp);
n = count; //因为count=0占一个有效节点，退出循环时，count值就是有效节点
for(i = 0; i //枚举完整的n/k块
for(int j = (i+1)*k - 1; j > i*k; j-- ){ //每块的第i个倒着输出，剩余最后一个节点
printf("%05d %d %05d\n",node[j].address,node[j].data,node[j-1].address);
}
printf("%05d %d ",node[i*k].address,node[i*k].data); //每块的最后一个节点的前两项数据
if(i 1) { //如果是非最后一块节点
printf("%05d\n",node[(i+2)*k-1].address);
}else{ //如果是最后一块节点
if(n % k == 0) printf("-1\n"); //刚好除整
else{ //如果最后一个节点不规则
printf("%05d\n",node[(i+1)*k].address);
for(i = n/k*k; i ){
printf("%05d %d ",node.address,node.data);
if(i 1){
printf("%05d\n",node[i+1].address);
}else{
printf("-1\n");
}// else
} // for(i)
}//else
}//else
} //for(i)
return 0;
}
转载于:https://www.cnblogs.com/wanghao-boke/p/8550233.html
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