本文主要是介绍PAT 1074 Reversing Linked List [静态链表] [无效结点]，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
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---------------------------------------这是题目和解题的分割线---------------------------------------
和这道题的思路差不多吧PAT 1097 Deduplication on a Linked List，但是我脑子大概是进浆糊了，最后两个测试点卡了好久，最后发现一个是没看清题目，是反转链表，即按照id倒序输出而不是按照data倒序= = 还有一个是反转的次数是有效结点/K，而不是N/K（明明注意到这一点但傻了忘了这个地方也要改 = = ）链表（静态）的题都要记得小心无效结点
大概思路是先按照单链表的顺序排序，再按照k分组排序。
#include<cstdio>
#include<algorithm>const int maxN = 100010;using namespace std;struct node
{int data;int address; //当前地址 int next;int id; //链表次序
}list[maxN];bool cmp1(node a,node b)
{return a.id<b.id;
}bool cmp2(node a,node b)
{return a.id>b.id;
}int main()
{int begin,n,k,i,j;scanf("%d%d%d",&begin,&n,&k);int address;for(i=0;i<n;i++){scanf("%d",&address);scanf("%d%d",&list[address].data,&list[address].next);list[address].address = address;}//初始化成超大的数，以便排序 for(i=0;i<maxN;i++)list[i].id = 2*maxN;int p = begin,order = 0;//遍历，并标上序号 while(p!=-1){list[p].id = order++;p = list[p].next;}//按单链表的顺序排序 sort(list,list+maxN,cmp1);//是order/k不是n/k 存在无效结点干扰的情况 int turn = order/k;for(i=0,j=0;j<turn;i=i+k,j++)sort(list+i,list+i+k,cmp2);for(i=0;i<order-1;i++)printf("%05d %d %05d\n",list[i].address,list[i].data,list[i+1].address);printf("%05d %d -1\n",list[order-1].address,list[order-1].data);return 0;
}换了个方法试试，如下：
#include<cstdio>
#include<algorithm>using namespace std;struct node
{int address;int data;int next;int id;
}list[100010],print[100010];bool cmp(node a,node b)
{return a.id>b.id;
}int main()
{int begin,n,k,i,cnt = 0;scanf("%d%d%d",&begin,&n,&k);for(i=0;i<n;i++){int a,b,c;scanf("%d%d%d",&a,&b,&c);list[a].address = a;list[a].data = b;list[a].next = c;}//将正确顺序存在另一个结构体中 while(begin!=-1){print[cnt++] = list[begin];print[cnt-1].id = cnt-1;begin = list[begin].next;}int t = 0;//需要reverse时也可以通过设置id，再排序来实现 //注意排序和输出都是cnt不是n，因为题目存在无效结点 for(i=0;i<cnt/k;i++){sort(print+t*k,print+t*k+k,cmp);t++;}for(i=0;i<cnt;i++){printf("%05d %d",print[i].address,print[i].data);if(i!=cnt-1) printf(" %05d\n",print[i+1].address);else printf(" -1\n");}return 0;
}
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