02-线性结构3 Reversing Linked List（25 point(s)） 【链表】

02-线性结构3 Reversing Linked List（25 point(s)）

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 (≤10​5​​) 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

思路

用 STL 模拟 链表

然后 要注意

可能会有 多余结点 不在链表上面

AC代码

#include <cstdio>
#include <cstring>
#include <ctype.h>
#include <cstdlib>
#include <cmath>
#include <climits>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <deque>
#include <vector>
#include <queue>
#include <string>
#include <map>
#include <stack>
#include <set>
#include <numeric>
#include <sstream>
#include <iomanip>
#include <limits>

#define CLR(a) memset(a, 0, sizeof(a))
#define pb push_back

using namespace std;
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
typedef pair <int, int> pii;
typedef pair <ll, ll> pll;
typedef pair<string, int> psi;
typedef pair<string, string> pss;

const double PI = 3.14159265358979323846264338327;
const double E = exp(1);
const double eps = 1e-30;

const int INF = 0x3f3f3f3f;
const int maxn = 1e5 + 5;
const int MOD = 1e9 + 7;

struct Node
{
    int add;
    int value;
    int next;
    void read()
    {
        scanf("%d%d%d", &add, &value, &next);
    }
}q[maxn];

vector <Node> v, ans;

map <int, Node> m;

void dfs(int add)
{
    v.pb(m[add]);
    if (m[add].next != -1)
        dfs(m[add].next);
}

int main()
{
    int start, n, k;
    scanf("%d%d%d", &start, &n, &k);
    for (int i = 0; i < n; i++)
    {
        q[i].read();
        m[q[i].add] = q[i];
    }
    dfs(start);
    n = v.size();
    int len = v.size() / k;
    for (int i = 1; i <= len; i++)
    {
        int j = i * k - 1;
        int m = k;
        while (m--)
            ans.pb(v[j--]);
    }
    for (int i = len * k; i < n; i++)
        ans.pb(v[i]);
    for (int i = 0; i < n - 1; i++)
        printf("%05d %d %05d\n", ans[i].add, ans[i].value, ans[i + 1].add);
    printf("%05d %d -1\n", ans[n - 1].add, ans[n - 1].value);
}