02-線性結構3 Reversing Linked List (25分)
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 (≤10e5 ) 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
參考鏈接
https://blog.csdn.net/liyuanyue2017/article/details/83269991
浙大數據結構習題筆記:02-線性結構3 Reversing Linked List (25分))
我寫了很久,題目弄錯了好幾次,以爲和我之前寫的題目一樣呢,結果怎麼改都不對,類似的題目之前見過,這個更復雜。
大佬寫的正確的答案:
#include<bits/stdc++.h>
using namespace std;
#define max 100000
int main() {
int data[max];
int next[max];
int list[max];
int firstAdd,n,k;
cin>>firstAdd>>n>>k;
for(int i=0; i<n; i++) {
int x,y,z;
cin>>x>>y>>z;
data[x]=y;
next[x]=z;
}
int sum=0;
while(firstAdd!=-1) {
list[sum++]=firstAdd;//記錄下了連續的地址
firstAdd=next[firstAdd];//下一個節點的地址
}
for(int i=0; i<sum-sum%k; i=i+k) {
for(int j=0; j<k/2; j++) {
int t=list[i+j];
list[i+j]=list[i+k-j-1];
list[i+k-j-1]=t;
}
}
for(int i=0; i<sum-1; i++)
printf("%05d %d %05d\n",list[i],data[list[i]],list[i+1]);
printf("%05d %d -1\n",list[sum-1],data[list[sum-1]]);
return 0;
}