pat 1074 Reversing Linked List (25分) 链表反转

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 (≤) which is the total number of nodes, and a positive K (≤) 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

解题思路：

这道题目，用一个vector，按照顺序把 节点依次存起来，输入是4 - > 1 - > 6 -> 3 ->5 -> 2  按照 1->2->3->4->5->6存放，再反转vector

#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<vector>
#include<algorithm>

using namespace std;

typedef struct Node{
	int data;
	int order;
	int next;
}Node;
Node nodes[100010];
int main(){
//	freopen("C:\\Users\\zzloyxt\\Desktop\\1.txt","r",stdin);
	
	int first,n,k;
	scanf("%d %d %d",&first,&n,&k);
	
	for(int i=0;i<n;i++){
		int data,order,next;
		scanf("%d %d %d",&data,&order,&next);
		nodes[data].data = data;
		nodes[data].order = order;
		nodes[data].next = next;		
	}
	vector<Node> V;
	
	for(int i=first;i!=-1;i = nodes[i].next){
		V.push_back(nodes[i]);
	}
	
	for(int i=0;i<=V.size()-k;i=i+k){
		reverse(V.begin()+i,V.begin()+i+k);
	}
	
	for(int i=0;i<V.size()-1;i++){
		printf("%05d %d %05d\n",V[i].data,V[i].order,V[i+1].data);
	}
	
	printf("%05d %d %d\n",V[V.size()-1].data, V[V.size()-1].order,-1);
	
	return 0;
}

 

注意这里的reverse()函数，传的是iterator，reverse(V.begin(), V.begin() + k)，最后一个单独输出