数据结构练习 02-线性结构2. Reversing Linked List (25)
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 提交了10次才完全正确,花了快5个小时。感觉自己实力好差。
#include<iostream>
#include <string>
using namespace std;
class linked{
public:
int data;
int next;
};
linked *list = new linked[];
int prehead;
int nexthead;
int Reverse(int head, int k){
int cnt = ,temp;
int news = head;
int olds = list[head].next;
while (cnt < k){
temp = list[olds].next;
list[olds].next = news;
news = olds;
olds = temp;
cnt++;
}
list[head].next = olds;
prehead = head;
nexthead = olds;
return news;
}
int main(){
int firstNode, address, next,N, k, data,head,preshead;
int sum = ;
cin >> firstNode >> N >> k;
for (int i = ; i < N; i++){
cin >> address >> data >> next;
list[address].data = data;
list[address].next = next;
}
int p = firstNode;
while (p != -){
sum++;
p = list[p].next;
}
head = Reverse(firstNode, k);
for (int i = ; i < sum / k; i++){
preshead = prehead;
list[preshead].next = Reverse(nexthead, k);
}
p = head;
while (p!= -){
printf("%05d ",p);
cout << list[p].data <<" ";
if (list[p].next != -)
printf("%05d\n", list[p].next);
else
cout << list[p].next << endl;
p = list[p].next;
} return ;
}
ac的图片看着真的很爽。
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