7-2 Play with Linked List (25分)

Given a singly linked list L​₁​​→L​_2​​→⋯→L​_n−1​​→L​_n​​ and an integer 1≤k<n, you are supposed to rearrange the links to obtain a list like L_​k​​→L_​n​​→L_{​k−1​​}→L_{​n−1​​}→⋯. For example, given L being 1→2→3→4→5→6 and k=4, you must output 4→6→3→5→2→1.

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 (≤10​⁵​​) which is the total number of nodes, and an integer 1≤k<n where n is the number of nodes in the linked list. 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 a positive integer no more than 10​^5​​, and Next is the position of the next node. It is guaranteed that there are at least two nodes on the list.

Output Specification:

For each case, output in order the resulting 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 68237
68237 6 33218
33218 3 99999
99999 5 12309
12309 2 00100
00100 1 -1

• 思路：
  链表题：直接模板处理，排序+输出

• T1 code:

#include <bits/stdc++.h>
using namespace std;
const int maxn = 100010;
struct Node{
	int add, data, next;
}node[maxn];
vector<int> preM, postM, ans;

int main(){
	int first, n, m, tmpAdd;
	scanf("%d %d %d", &first, &n, &m);
	for(int i = 0; i < n; ++i){
		scanf("%d", &tmpAdd);
		scanf("%d %d", &node[tmpAdd].data, &node[tmpAdd].next);
		node[tmpAdd].add = tmpAdd;
	}
	int cnt = m; 
	while(first != -1){
		if(cnt-- > 0){
			preM.push_back(node[first].add);
		} 
		else{
			postM.push_back(node[first].add);
		} 
		first = node[first].next;
	}
//	for(int i = 0; i < preM.size(); ++i)
//		cout  << preM[i] << " ";
	reverse(preM.begin(), preM.end());
	reverse(postM.begin(), postM.end());
	int i;
	for(i = 0; i < preM.size() && i < postM.size(); ++i){
		ans.push_back(preM[i]);
		ans.push_back(postM[i]);
	}
//	cout << i;
	if(i == preM.size()){
		for(int j = i; j < postM.size(); ++j)	ans.push_back(postM[j]);
	}else if(i == postM.size()){
		for(int j = i; j < preM.size(); ++j)	ans.push_back(preM[j]);	
	}
//	cout << node[0].data;
	int j;
	for(j = 0; j < ans.size()-1; ++j)
		printf("%05d %d %05d\n", ans[j], node[ans[j]].data, node[ans[j+1]].add);
	printf("%05d %d -1\n", ans[j], node[ans[j]].data);
	return 0;
}

• T2 code:

#include <bits/stdc++.h>
using namespace std;
const int maxn = 100010;
struct Node
{
    int data, nex;
}node[maxn];

void Print(const vector<int> & ans)
{
    if(ans.size() == 0)
    {
        printf("0 -1\n");
    }else
    {
        for(int i = 0; i < ans.size(); ++i)
        {
            if(i < ans.size()-1) printf("%05d %d %05d\n", ans[i], node[ans[i]].data, ans[i+1]);
            else printf("%05d %d -1\n", ans[i], node[ans[i]].data);
        }
    }
}
int main()
{
    int first, n, k;
    scanf("%d %d %d", &first, &n, &k);
    for(int i = 0; i < n; ++i)
    {
        int add;
        scanf("%d", &add);
        scanf("%d %d", &node[add].data, &node[add].nex);
    }
    int head = first, idex = k;
    vector<int> ans[3];
    while(head != -1)
    {
        if(idex-- > 0){
            ans[1].push_back(head);
        }else
        {
            ans[2].push_back(head);
        }
        head = node[head].nex;
    }
    int i, j, len1 = ans[1].size(), len2 = ans[2].size();
    for(i = 0, j = 0; i < len1 && j < len2; ++i, ++j)
    {
        ans[0].push_back(ans[1][len1 - 1 - i]);
        ans[0].push_back(ans[2][len2 - 1 - j]);
    }
    while(i < ans[1].size())
    {
        ans[0].push_back(ans[1][len1 - 1 - i]);
        i++;
    }
    while(j < ans[2].size())
    {
        ans[0].push_back(ans[2][len2 - 1 - j]);
        j++;
    }
    Print(ans[0]);
    return 0;
}