PAT (Advanced Level) Practise 1074 Reversing Linked List (25)
2017-07-07 16:39
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1074. Reversing Linked List (25)
时间限制400 ms
内存限制
65536 kB
代码长度限制
16000 B
判题程序
Standard
作者
CHEN, Yue
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
题意:给你一个链表的头,让你每隔k个节点倒置一下节点
解题思路:模拟,这题好像比较坑的一点是不一定是一个链表
#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
using namespace std;
#define LL long long
const int INF = 0x3f3f3f3f;
struct node
{
int val,nt;
}a[100009];
int s,n,k,x,ans[100009],cnt;
int main()
{
while(~scanf("%d%d%d",&s,&n,&k))
{
for(int i=1;i<=n;i++) scanf("%d",&x),scanf("%d%d",&a[x].val,&a[x].nt);
cnt=0;
for(int i=s;~i;i=a[i].nt) ans[cnt++]=i;
for(int i=0;i+k<=cnt;i+=k)
{
for(int j=0;j<=k;j++)
{
if(i+j>=i+k-j-1) break;
swap(ans[i+j],ans[i+k-j-1]);
}
}
for(int i=0;i<cnt;i++)
{
printf("%05d %d ",ans[i],a[ans[i]].val);
if(i==cnt-1) printf("-1\n");
else printf("%05d\n",ans[i+1]);
}
}
return 0;
}
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